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ClassdJLLQi_ 
Book .jLJaMto 



/ 

ROBINSON'S MATHEMATICAL SERIES. 1 



THE 

PROGKESSIVE 

HIGHER ARITHMETIC, 

FOB 

SCHOOLS, ACADEMIES, AND MERCANTILE COLLEGES. 

COMBINING THE 

ANALYTIC AND SYNTHETIC METHODS; 



AND FORMING A COMPLETE TREATISE ON ARITHMETICAL 

SCIENCE, AND ITS COMMERCIAL AND 

BUSINESS APPLICATIONS. 

BY 

HORATIO N. ROBINSON, LL. D., 

AUTHOR OF WORKS ON ALGEBRA, GEOMETRY AND TRIGONOMETRY, SURVEYING AND 
NAVIGATION, CONIC SECTIONS, CALCULUS, ASTRONOMY, ETC 



►-•-+- 



-NEW YORK: 

IVISON, PHINNEY CO., 48 AND 50 WALKER ST. 

CHICAGO : S. C. GRIGGS <fc CO., 39 AND 41 LAKE ST. 

BOSTON I BROWN k TAGGARD. PHILADELPHIA : SOWER, BARNES k CO., AND 

J. B. LIPPINCOTT k CO. CINCINNATI : MOORE, WILSTACH, KEYS k CO. 

SAVANNAH: J. M. COOPER k CO. ST. LOUIS: KEITH k WOODS. 

NEW ORLEANS : E. R. STEVENS k CO. DETROIT : RAYMOND 

k LAPHAM. BALTIMORE I CUSHLNG k BAILEY. 

1860. 



Q/j/o3 

Robinson's Complete Mathematical Course". 
ROBINSON'S SYSTEM* OP MATHEMATICS, 

Recently revised and enlarged, is now the most extensive, complete, prac- 
tical and scientific Mathematical Series published in this country. 
*—*—•> -*-** 

1. Robinson's Progressive Primary Arithmetic. Illustrated. $0 15 

2. Robinson's Progressive Intellectual Arithmetic, for ad- 
vanced Classes, with an Original and Comprehensive System 
of Analysis 25 

3. Robinson's Progressive Practical Arithmetic : a complete 
work for Common Schools and Academies 56 

4. Key to Robinson's Progressive Practical Arithmetic. . 50 

5. Robinson's Progressive Higher Arithmetic. . . . 75 

6. Key to Robinson's Progressive Higher Arithmetic. . . 75 

7. Robinson's New Elementary Algebra: a clear and simple 
Treatise for Beginners 75 

8. Key to Robinson's New Elementary Algebra. . . . 75 

9. Robinson's University Algebra : a full and complete Trea- 
tise for Academies and Colleges 1 25 

10. Key to Robinson's University Algebra; separate. . . 1 00 

11. Robinson's Geometry and Trigonometry; with applications 
to practical examples. (New.) 1 50 

12. Robinson's Surveying and Navigation ; combining theory 
with practice 1 50 

13. Robinson's Analytical Geometry and Conic Sections ; made 
clear and comprehensive to common minds 1 50 

14. Robinson's Differential and Integral Calculus; a full and 
complete Treatise 1 50 

15. Robinson's Elementary Astronomy; designed to teach the 
first principles of this Science. 75 

16. Robinson's University Astronomy; for advanced classes 
in Academies and Colleges 1 75 

17. Robinson's Concise Mathematical Operations : a book of 
reference for the Teacher, embracing the gems of Mathemati- 
cal Science 2 25 

18. Key to Robinson's University Algebra, Geometry, Sur- 
veying, and Calculus ; in 1 vol. . . . . . . 1 50 



Entered, according to Act of Congress, in the year 1SG0, by 

D. W FISH AND J. II. FRENCH, 

i" tli.- Clerl istrict Court of the United States for the Northern DistriJ 

of New York. 

— "il 



JUUN FAG AN, 8TEREOTYPER, PHILADELPHIA. 

1 -V *l \ 1- 



PREFACE. 



This work is intended to complete a well graded and 
progressive series of Arithmetics, and to furnish to ad- 
vanced students a more full and comprehensive text-book 
on the Science of Numbers than has before been published ; 
a work that shall embrace those subjects necessary to give 
the pupil a thoroughly practical and scientific arithmetical 
education, either for the farm, the workshop, or a profes- 
sion, or for the more difficult operations of the counting- 
room and of mercantile and commercial life. 

There are two general methods of presenting the ele- 
ments of arithmetical science, the Synthetic and the Ana- 
lytic. Comparison enters into every operation, from the 
simplest combination of numbers to the most complicated 
problems in the Higher Mathematics. Analysis first 
generalizes a subject and then develops the particulars of 
which it consists; Synthesis first presents particulars, 
from which, by easy and progressive steps, the pupil is led 
to a general and comprehensive view of the subject/ 
Analysis separates truths and properties into their ele- 
ments or first principles ; Synthesis constructs general 
principles from particular cases. Analysis appeals more 
(to the reason, and cultivates the desire to search for first 
principles, and to understand the reason for every process 
rather than to know the rule. Hence, the leading method in 
an elementary course of instruction should be the Synthetic, 
while in an advanced course it should be the Analytic. 

The following characteristics of a first class text-book 
will be obvious to all who examine this work : the typogra- 

(iii) 



J v PREFACE. 

phy and mechanical execution; the philosophical and 
scientific arrangement of the subjects ; clear and concise 
definitions ; fidl and rigid analyses ; exact and compre- 
hensive rules; brief and accurate methods of operation: 
the wide range of subjects and the large number and prac- 
tical character of the examples — in a word, scientific ac- 
curacy combined with practical utility, throughout the 
entire work. 

Much labor and attention have been devoted to obtain- 
ing correct and adequate information pertaining to mer- 
cantile and commercial transactions, and the Government 
Standard units of measures, weights, and money. The 
counting-room, the bank, the insurance and broker's office, 
the navy and ship-yard, the manufactory, the wharves, the 
custom-house, and the mint, have all been visited, and the 
most reliable statistics and the latest statutes have been 
consulted, for the purpose of securing entire accuracy in 
those parts of this work which relate to these subjects 
and departments. As the result of this thorough investi- 
gation, many statements found in most other arithmetics 
of a similar grade will not agree with the facts presented in 
this work, and simply because the statements in these 
other books have been copied from older works, while laws 
and customs have undergone great changes since the older 
works were written. 

New material and new methods will be found in the seve- 
ral subjects throughout the entire work. Considerable pro- 
minence has been given to Percentage and its numerous ap- 
plications, especially to Stocks, Insurance, Interest, Aver- 
aging Accounts, Domestic and Foreign Exchange, and seve- 
ral other subjects necessary to qualify students to become 
good accountants or commercial business men. And while 
this work may embrace many subjects not necessary to the 



PREFACE. v 

course usually prescribed in Mercantile and Commercial 
Colleges, yet those subjects requisite to make good account- 
ants, and which have been taught orally in that class of 
institutions from want of a suitable text-book, are fully dis- 
cussed and practically applied in this work ; and it is there- 
fore believed to be better adapted to the wants of Mercan- 
tile Colleges than any similar work yet published. And 
while it is due, it is also proper here to state that J. C. 
Porter, A. M., an experienced and successful teacher of 
Mathematics in this State, and formerly professor of Com- 
mercial Arithmetic, in Iron City Commercial College, Pitts- 
burgh, Penn., has rendered Valuable aid in the preparation 
of the above-named subjects, and of other portions of the 
work. He is likewise the author of the Factor Table on 
pages 72 and 73, and of the new and valuable improvement 
in the method of Cube Eoot. 

Teachers entertain various views relative to having the 
answers to problems and examples inserted in a text-book. 
Some desire the answers placed immediately after the ex- 
amples ; others wish them placed together in the back part 
of the book; and still others desire them omitted alto-, 
gether. All these methods have their advantages and their 
disadvantages. 

If all the answers are given, there is danger that the 
pupil will become careless, and not depend enough upon the 
accuracy of his own computations. Hence he is liable to 
neglect the cultivation of those habits of patient investiga- 
tion and self-reliance which would result from his being 
obliged to test the truth and accuracy of his own processes 
by proof, — the only test he will have to depend upon in all 
the computations in real business transactions in after life. 
Besides, the work of proving the correctness of a result is 
often of quite as much value to the pupil as the work of 
1* 



y - PREFACE. 

performing the operation ; as the proof may render simple 
and clear some part or the whole of an operation that was 
before complicated and obscure. 

If answers ara placed in the back part of the book, the 
pupil will at once refer to them whenever he is in any doubt 
or difficulty in performing an operation. Hence the object 
aimed at is not accomplished by placing the answers to- 
gether in this manner. 

Again, if all the answers are omitted, the pupil may be- 
come involved in doubt and uncertainty, and acquire a 
distaste for the study ; and from this discouragement, sub- 
sequently make but limited advancement in Mathematical 
Science. 

In order, therefore, that pupils may receive the advan- 
tages of both methods, the answers to nearly one half 
of the examples in this book are omitted. They will be 
found, together with full and clear solutions of all the 
examples, in a Key to this work, which has been prepared 
for the use of teachers and private learners. 

Many valuable hints and suggestions which have been 
received from teachers and friends of education, have 
been incorporated into this work. The author desires to 
make especial acknowledgment of the valuable services 
rendered in the preparation of this work by Mr. D. "W. Fish, 
of Eochester, K Y., and J. H. French, LL. D., of Syracuse, 
N. Y. These gentlemen have had long and successful 
experience as teachers, and an intimate acquaintance with 
the plans and operations of some of the best schools in the 
country. It is due to them to state that they are in fact 
the authors of the Practical and Higher Arithmetics of this 
series. 

August 1, 1860. 



CONTENTS. 



PAGE 

Definitions 11 

Signs 13 

Axioms 14 

Notation and Numeration 15 

SIMPLE NUMBERS. 

Addition 23 

Adding two or more columns at one operation 27 

Subtraction 30 

Two or more subtrahends 33 

Multiplication 35 

Powers of Numbers 39 

Continued Multiplication 40 

Contractions in Multiplication 41 

Division 47 

Abbreviated Long Division 50 

Successive Division 55 

Contractions in Division 55 

General Problems in Simple Numbers 61 

PROPERTIES OF NUMBERS. 

Exact Divisors 65 

Prime Numbers 68 

Table of Prime Numbers * 70 

Factoring . 70 

Factor Table 72 

Greatest Common Divisor 76 

Least Common Multiple 82 

Cancellation 86 

FRACTIONS. 

Definitions, Notation and Numeration 89 

Reduction 92 

Addition 99 

(vii) 



yiii CONTENTS. 

PAGE 

Subtraction 101 

Theory of Multiplication and Division 103 

Multiplication 104 

Division 107 

Greatest Common Divisor Ill 

Least Common Multiple 112 

DECIMALS. 

Notation and Numeration 117 

Beduction 121 

Addition 124 

Subtraction 126 

Multiplication t 127 

Contracted Multiplication 128 

Division 132 

Contracted Division 134 

Circulating Decimals 136 , 

Reduction of Circulating Decimals 139 

Addition and Subtraction of Circulating Decimals 142 

Multiplication and Division of Circulating Decimals 144 

UNITED STATES MONEY. 

Notation and Numeration 145 

Reduction 147 

Operations 147 

Problems 150 

Ledger Accounts 153 

Accounts and Bills 153 

Continued Fractions 161 

COMPOUND NUMBERS. 

Measures of Extension 164 

Measures of Capacity ** 170 

Measures of Weight 171 

Measure of Time 175 

Measure of Angles 177 

Miscellaneous Tables 178 

Government Standards of Measures and Weights 179 

English Measures and Weights 182 

French Measures and Weights 184 

Money and Currencies * 187 

Reduction 192 

Reduction Descending 192 



CONTENTS. ix 

PAGE 

Reduction Ascending.... 199 

Addition 206 

Subtraction 209 

Multiplication 214 

Division 216 

Longitude and Time 218 

DUODECIMALS. 

Addition and Subtraction 227 

Multiplication 228 

Division 230 

SHORT METHODS. 

For Subtraction 232 

For Multiplication 233 

For Division 241 

RATIO. 243 

PROPORTION. 247 

Cause and Effect 249 

Simple Proportion... 249 

Compound Proportion 253 

PERCENTAGE. 

Notation 259 

General Problems 260 

Applications 268 

Commission 268 

Stocks 272 

Stock-jobbing 273 

Instalments, Assessments, and Dividends 276 

Stock Investments 279 

Profit and Loss 283 

Insurance 287 

Fire and Marine Insurance 287 

Life Insurance 289 

Life Table 291 

Endowment Assurance Table 292 

Taxes 294 

(Jsneral Average 297 

Custom House Business 299 

Simple Interest 303 



X 



CONTENTS. 



PAGE 

Partial Payments or Indorsements 310 

Savings Banks Accounts 315 

Compound Interest 317 

Compound Interest Table 319 

Problems in Interest » 320 

Discount 324 

Banking 326 

Exchange 333 

Direct Exchange 335 

Table of Foreign Coins and Money 338 

Arbitrated Exchange 344 

Equation of Payments 348 

Simple Equations 348 

Compound Equations 353 

Account Sales 357 

Settlement of Accounts Current 359 

Partnership 360 

ALLIGATION. 366 

INVOLUTION. 375 

EVOLUTION. 377 

Square Root 378 

Contracted Method 382 

Cube Root 383 

Contracted Method 388 

SERIES. 394 

Arithmetical Progression 395 

Geometrical Progression 397 

Compound Interest by Geometrical Progression 401 

Annuities 402 

Annuities at Simple Interest 402 

Annuities at Compound Interest ,. 404 

MENSURATION. 

Lines and Surfaces 406 

Solids 411 

Miscellaneous Examples 415 



HIGHER ARITHMETIC. 



DEFINITIONS. 

1. Quantity is any thing that can be increased, diminished, or 
measured ; as distance, space, weight, motion, time. 

2. A Unit is one, a single thing, or a definite quantity. 

3. A Number is a unit, or a collection of units. 

4. The Unit of a Number is one of the collection constituting 
the number. Thus, the unit of 34 is 1; of 34 days is 1 day. 

5. An Abstract Number is a number used without reference 
to any particular thing or quantity; as 3, 24, 756. 

6. A Concrete Number is a number used with reference to 
some particular thing or quantity; as 21 hours, 4 cents, 230 miles. 

7. Unity is the unit of an abstract number. 

8. The Denomination is the name of the unit of a concrete 
number. 

9. A Simple Number is either an abstract number, or a con- 
crete number of but one denomination; as 48, 52 pounds, 36 days. 

10. A Compound Number is a concrete number expressed in 
two or more denominations ; as, 4 bushels 3 pecks, 8 rods 4 yards 
2 feet 3 inches. 

11. An Integral Number, or Integer, is a number which ex- 
presses whole things; as 5, 12 dollars, 17 men. 

12. A Fractional Number, or Fraction, is a number which 
expresses equal parts of a whole thing or quantity; as J, f of a 
pound, T 5 ^ of a bushel. 

13. Like Numbers have the same kind of unit, or express the 
same kind of quantity. Thus, 74 and 16 are like numbers; so 
are 74 pounds, 16 pounds, and 12 pounds; also, 4 weeks 3 days, and 
16 minutes 20 seconds, both being used to express units of time. 

14. Unlike Numbers have different kinds of units, or are used 

(") 



12 SIMPLE NUMBERS. 

to express different kinds of quantity. Thus, 36 miles, and 15 
days ; 5 hours 36 minutes, and 7 bushels 3 pecks. 

15. A Power is the product arising from multiplying a number 
by itself, or repeating it any number of times as a factor. 

1G. A Boot is a factor repeated to produce a power. 

17. A Scale is the order of progression on which any system 
of notation is founded. Scales are uniform and varying. 

18. A Uniform Scale is one in which the order of progression 
is the same throughout the entire succession of units. 

19. A Varying Scale is one in which the order of progression 
is not the same throughout the entire succession of units. 

20. A Decimal Scale is one in which the order of progression 
is uniformly ten. 

21. Mathematics is the science of quantity. 

The two fundamental branches of Mathematics are Geometry 
and Arithmetic. Geometry considers quantity with reference to 
positions, form, and extension. Arithmetic considers quantity as 
an assemblage of definite portions, and treats only of those condi- 
tions and attributes which may be investigated and expressed by 
numbers. Hence, 

22. Arithmetic is the Science of numbers, and the Art of 

computation. 

Note. — When Arithmetic treats of operations on abstract numbers it is a sci- 
ence, and is then called Pure Arithmetic. When it treats of operations on con- 
crete numbers it is an art, and is then called Applied Arithmetic. Pure and 
Applied Arithmetic are also called Theoretical and Practical Arithmetic. 

23. A Demonstration is a process of reasoning by which a 
truth or principle is established. 

24. An Operation is a process in which figures are employed 
to make a computation, or obtain some arithmetical result. 

25. A Problem is a question requiring an operation. 

26. A Rule is a prescribed method of performing an operation. 

27. Analysis, in arithmetic, is the process of investigating 
principles, and solving problems, independently of set rules. 

28. The Five Fundamental Operations of Arithmetic are, 
Notation and Numeration, Addition, Subtraction, Multiplication, 
and Division. 



DEFINITIONS. 13 



SIGNS. 



29. A Sign is a character indicating the relation of numbers, 
or an operation to be performed. 

30. The Sign of Numeration is the comma (,). It indicates 
that the figures set off by it express units of the same general name, 
and are to be read together, as thousands, millions, billions, etc. 

31. The Decimal Sign is the period (.). It indicates that 
the number after it is a decimal. 

32. The Sign of Addition is the perpendicular cross, +, called 
plus. It indicates that the numbers connected by it are to be 
added; as 3 + 5 + 7, read 3 plus 5 plus 7. 

33. The Sign of Subtraction is a short horizontal line, — , 
called, minus. It indicates that the number after it is to be sub- 
tracted from the number before it; as 12 — 7, read 12 minus 7. 

34. The Sign of Multiplication is the oblique cross, x . It 
indicates that the numbers connected by it are to be multiplied 
together ; as 5 x 3 x 9, read 5 multiplied by 3 multiplied by 9. 

35. The Sign of Division is a short horizontal line, with a 
point above and one below, H-. It indicates that the number 
before it is to be divided by the number after it; as 18 -+- 6, read 
18 divided by 6. 

Division is also expressed by writing the dividend above, and 
the divisor below, a short horizontal line. Thus, l f, read 18 
divided by 6. 

36. The Sign of Equality is two short, parallel, horizontal 
lines, =. It indicates that the numbers, or combinations of 
numbers, connected by it are equal; as 4 + 8 = 15 — 3, read 4 
plus 8 is equal to 15 minus 3. Expressions connected by the 
sign of equality are called equations. 

37. The Sign of Aggregation is a parenthesis, ( ). It indi- 
cates that the numbers included within it are to be considered 
together, and subjected to the same operation. Thus, (8 + 4) x 5 
indicates that both 8 and 4, or their sum, is to be multiplied by 5. 

A vinc ulum or bar, , has the same signification. Thus, 

7x9--3 = 21. 

2 



14 SIMPLE NUMBERS. 

38. The Sign of Ratio is two points, : . Thus, 7 : 4 is read, 
the ratio of 7 to 4. 

39. The Sign of Proportion is four points, : : . Thus, 

3 : 6 : : 4 : 8, is read, 3 is to 6 as 4 is to 8. 

40. The Sign of Involution is a number written above, and a 
little to the right, of another number. It indicates the power to 
which the latter is to be raised. Thus, 12 3 indicates that 12 is 
to be taken 3 times as a factor; the expression is equivalent to 
12 x 12 X 12. The number expressing the sign of involution is 
called the Index or Exponent. 

41. The Sign of Evolution, v% is a modification of the letter r. 
It indicates that some root of the number after it is to be extracted. 
Thus, v/25 indicates that the square root of 25 is to be extracted; 
•\K64 indicates that the cube root of 64 is to be extracted. 

AXIOMS. 

42. An Axiom is a self-evident truth. The principal axioms 
required in arithmetical investigations are the following : 

1. If the same quantity or equal quantities be added to equal 
quantities, the sums will be equal. 

2. If the same quantity or equal quantities be subtracted from 
equal quantities, the remainders will be equal. 

3. If equal quantities be multiplied by the same number, the 
products will be equal. 

4. If equal quantities be divided by the same number, the quo- 
tients will be equal. 

5. If the same number be added to a quantity and subtracted 
from the sum, the remainder will be that quantity. 

6. If a quantity be multiplied by a number and the product 
divided by the same number, the quotient will be that quantity. 

7. Quantities which are respectively equal to any other quantity 
are equal to each other. 

8. Like powers or like roots of equal quantities are equal. 

9 The whole of any quantity is greater than any of its parts. 
10. The whole of any quantity is equal to the sum of all its 
parts. 



NOTATION AND NUMERATION. J5 



NOTATION AND NUMERATION. 

43. Notation is a method of writing or expressing numbers by 
characters; and, 

44. Numeration is a method of reading numbers expressed 
by characters. 

45. Two systems of notation are in general use — the Roman 
and the Arabic. 

Note. — The Roman Notation is supposed to have been first used by the 
Romans : hence its name. The Arabic Notation was first introduced into Europe 
by the Moors or Arabs, who conquered and held possession of Spain during the 
11th century. It received the attention of scientific men in Italy at the begin- 
ning of the 13th century, and was soon afterward adopted in most European 
countries. Formerly it was supposed to be an invention of the Arabs; but 
investigations have shown that the Arabs adopted it from the Hindoos, among 
whom it has been in use more than 2000 years. From this undoubted origin it 
is sometimes called the Indian Notation. 

THE ROMAN NOTATION. 

46. Employs seven capital letters to express numbers. Thus, 
Letters, I V X L C D M 

Values, one, five, ten, fifty, , OT \ e . , fiv , e . .~ one . 

A ' — ' J ' hundred, hundred, thousand. 

47. The Roman notation is founded upon five principles, as 
follows : 

1st. Repeating a letter repeats its value. Thus, II represents 
two, XX twenty, CCC three hundred. 

2d. If a letter of any value be placed after one of greater value, 
its value is to be united to that of the greater. Thus, XI repre- 
sents eleven, LX sixty, DC six hundred. 

3d. If a letter of any value be placed be/ore one of greater value, 
its value is to be taken from that of the greater. Thus, IX repre- 
sents nine, XL forty, CD four hundred. 

4th. If a letter of any value be placed betioeen two letters, each 
of greater value, its value is to be taken from the united value of 
the other two. Thus, XIV represents fourteen, XXIX twenty- 
nine, XCIV ninety-four. 

5th. A bar or dash placed over a letter increases its value one 
thousand fold. Thus, V signifies five, and V five thousand ; L 
fifty, and L fifty thousand. 



16 



SIMPLE NUMBERS. 



TABLE 


OF ROMAN NOTATION. 


I is One. 


XX is Twenty. 


II " Two. 


XXI " Twenty-one. 


Ill " Three. 


XXX " Thirty. 


IV " Four. 


XL " Forty. 


V " Five. 


L " Fifty. 


VI " Six. 


LX " Sixty. 


VII " Seven. 


LXX " Seventy. 


VIII " Eight. 


LXXX " Eighty. 


IX " Nine. 


XC " Ninety. 


X " Ten. 


C " One hundred. 


XI " Eleven. 


CC " Two hundred. 


XII " Twelve. 


D " Five hundred. 


XIII " Thirteen. 


DC " Six hundred. 


XIV " Fourteen. 


M " One thousand. [dred. 


XV " Fifteen. 


MC " One thousand one hun- 


XVI " Sixteen. 


MM " Two thousand. 


XVII " Seventeen. 


X " Ten thousand. 


XVIII " Eighteen. 


C " One hundred thousand. 


XIX " Nineteen. 


■ M " One million. 



Notes. — 1. Though the letters used in the above table have been employed 
as the Roman numerals for many centuries, the marks or characters used origi- 
nally in this notation are as follows : 

Modern numerals, I V X L C D M 

Primitive characters, I V X L C N M 

2. The system of Roman Notation is not well adapted to the purposes of nu- 
merical calculation ; it is principally confined to the numbering of chapters and 
sections of books, public documents, etc. 



EXAMPLES FOR PRACTICE. 

Express the following numbers by the Koman notation: 

1. Fourteen. 6. Fifty-one. 

2. Nineteen. 7. Eighty-eight. 
3\ Tweifty-four. 8. Seventy-three. 

4. Thirty-nine. 9. Ninety-five. 

5. Forty-six. 10. One hundred one. 

11. Five hundred fifty-five. 

12. Seven hundred ninety-eight. 

13. One thousand three. 

14. Twenty thousand eight hundred forty-five. 



NOTATION AND NUMERATION. 



17 



THE ARABIC NOTATION 

48. Employs ten characters or figures to express numbers. 
Thus, 

Figures, 123456789 

Names and ) nau § ht one > tw0 > three > four > five > six > seven, eight, nine, 
values. \ dp °^ 

49. The cipher, or first character, is called naught, because it 
has no value of its own. It is otherwise termed nothing, and zero. 
The other nine characters are called significant figures, because 
each has a value of its own. They are also called digits, a word 
derived from the Latin term digitus, which signifies finger. 

50. The ten Arabic characters are the Alphabet of Arithmetic. 
Used independently, they can express only the nine numbers that 
correspond to the names of the nine digits. But when combined 
according to certain principles, they serve to express all numbers. 

51. The notation of all numbers by the ten figures is accom- 
plished by the formation of a series of units of different values, to 
which the digits may b& successively applied. First, ten simple 
units are considered together, and treated as a single superior 
unit ; then, a collection of ten of these new units is taken as a still 
higher unit; and so on, indefinitely. A regular series of units, in 
ascending orders, is thus formed, as shown in the following 

TABLE OF UNITS. 

Primary units are called units of the first order. 

Ten units of the first order make 1 unit " " second " 
Ten " " " second " " 1 " " " third " 
Ten " " " third " " 1 " " " fourth " 
etc., etc. etc., etc. 

52. The various orders of units, when expressed by figures, 
are distinguished from each other by their location, or the place 
they occupy in a horizontal row of figures. Units of the first order 
are written at the right hand ; units of the second order occupy 
the second place ; units of the third order the third place ; and so 
on, counting from right to left, as shown on the following page : 

2* B 



18 



SIMPLE NUMBERS. 



ooooooooo 

53. In this notation we observe — 

1st. That a figure written in the place of any order, expresses 
as many units of that order as is denoted by the name of the figure 
used. Thus, 436 expresses 4 units of the 3d order, 3 units of the 
2d order, and 6 units of the 1st order. 

2d. The cipher, having no value of its own, is used to fill the 
places of vacant orders, and thus preserve the relative positions of 
the significant figures. Thus, in 50, the cipher shows the absence 
of simple units, and at the same time gives to the figure 5 the 
local value of the second order of units. 

54. Since the number expressed by any figure depends upon 
the place it occupies, it follows that figures have two values, 
Simple and Local. 

55. The Simple Value of a figure is its value when taken 
alone ; thus, 4, 7, 2. 

56. The Local Value of a figure is its value when used with 

another figure or figures in the same number. Thus, in 325, the 

local value of the 3 is 300, of the 2 is 20, and of the 5 is 5 units. 

Note. — When a figure occupies units' place, its simple and local values are 
the same. 

57. The leading principles upon which the Arabic notation 
is founded are embraced in the following 

GENERAL LAWS. 

I. All numbers are expressed by applying the ten figures to dif- 
ferent orders of units. 

II. The different orders of units increase from right to left, and 
decrease from left to right, in a tenfold ratio. 

Til. Every removal of a figure one place to the left, increases its 
local value tenfold; and every removal of afigune one place to the 
right, diminishes its local value tenfold. 



NOTATION A^D NUMERATION. 19 

58. In numerating, or expressing numbers verbally, the various 
orders of units have the following names : 

ORDERS. NAMES. 

1st order is called Units. 

2d order " " Tens. 

3d order " " Hundreds. 

4th order " " Thousands. 

5th order " " Tens of thousands. 

6th order " " Hundreds of thousands. 

7th order " " Millions. 

8th order " " Tens of millions. 

9th order " " Hundreds of millions, 

etc., etc. etc., etc. 

59. This method of numerating, or naming, groups the suc- 
cessive orders into periods of three figures each, there being three 
orders of thousands, three orders of millions, and so on in all 
higher orders. These periods are commonly separated by commas, 
as in the following table, which gives the names of the orders 
and periods to the twenty-seventh place. 




9 8,765,43 2,1 09,8 7 6,5 5 6,7 8 9,0 1 2,3 4 5 



ninth eighth seventh sixth fifth fourth third second first 
period, period, period, period, period, period, period, period, period. 

Note. — This is the French method of numerating, and is the one in general 
use in this country. The English numerate by periods of six figures each. 

©O. The names of the periods are derived from the Latin 
numerals. The twenty-two given on the following page extend 
the numeration table to the sixty-sixth place or order, inclusive. 



20 



SIMPLE NUMBERS. 



PERIODS. 


NAMES. 


PERIODS. 


NAMES. 


1st 


Units. 


12th 


Decillions. 


2d 


Thousands. 


13th 


Undecillions. 


3d 


Millions. 


14th 


Duodecillions. 


4th 


Billions. 


15th 


Tredecillions. 


5th 


Trillions. 


16th 


Quatuordecillions 


6th 


Quadrillions. 


17th 


Quindecillions. 


Tth 


Quintillions. 


18th 


Sexdecillions. 


8th 


Sextillions. 


19th 


Septendecillions. 


9th 


Septillions. 


20th 


Octodecillions. 


10th 


Octillions. 


21st 


Novendecillions. 


11th 


Nonillions. 


22d 


Vigintillions. 



01. From this analysis of the principles of Notation and Nume- 
ration, we derive the following rules : 

RULE FOR NOTATION. 

I. Beginning at the left hand, write the figures belonging to the 
highest period. 

II. Write the hundreds, tens, and units of each successive period 
in their order, placing a cipher wherever an order of units is 
omitted. 

RULE FOR NUMERATION. 

I. Separate the number into periods of three figures each, com- 
mencing at the right hand. 

II. Beginning at the left hand, read each period separately, and 
give the name to each period, except the last, or period of units. 

Note. — Omit and in reading the orders of units and periods of a number. 



EXAMPLES FOR PRACTICE. 

Write and read the following numbers : — 

1 . One unit of the 3d order, two of the 2d, five of the 1st. 

Ans. 125 ; read, one hundred twenty-five. 

2. Two units of the 5th order, four of the 4th, five of the 2d, 
six of the 1st. Ans. 24056 ; read, twenty-four thousand fifty-six. 

3. Seven units of the 4th order, five of the 3d, three of the 2d, 
ei^ht of the 1st 



NOTATION AND NUMERATION. 21 

4. Nine units of the 4th order, two of the 3d, four of the 1st. 

5. Five units of the 4th order, eight of the 2d. 

6. Five units of the 5th order, one of the 3d, eight of the 1st. 

7. Three units of the 5th order, six of the 4th, four of the 3d, 
seven of the 1st. 

8. Two units of the 6th order, four of the 5th, nine of the 4th, 
three of the 3d, five of the 1st. 

9. Three units of the 8th order, five of the 7th, four of the 6th, 
three of the 5th, eight of the 4th, five of the 3d, eight of the 2d, 
seven of the 1st. 

10. Three units of the 9th order, eight of the 7th, four of the 
6th, six of the 5th, nine of the 1st. 

11. Five units of the 12th order, three of the 11th, six of the 
10th. 

12. Four units of the 12th order, five of the 10th, eight of the 
5th, nine of the 4th, four of the 3d. 

13. Three units of the 15th order, six of the 14th, five of the 
13th, three of the 9th, six of the 8th, five of the 7th, three of the 
3d, six of the 2d, five of the 1st. 

14. Five units of the 18th order, three of the 17th, six of the 
16th, four of the 15th, seven of the 14th, eight of the 13th, four 
of the 12th, five of the 11th, six of the 10th, seven of the 9th, 
eight of the 8th, nine of the 7th, five ot the 6th, six of the 5th, 
three of the 4th, two of the 3d, four of the 2d, eight of the 1st. 

15. Two units of the 20th order, seven of the 19th, four of the 
18th, eight of the 13th, five of the 6th, five of the 5th, five of the 
4th, nine of the 1st. 

Write the following numbers in figures • 

16. Forty-eight. 

17. One hundred sixty-four. 

18. Forty-eight thousand seven hundred eighty-nine. 

19. Five hundred thirty-six million three hundred forty-seven 
thousand nine hundred seventy-two. 

20. Ninety-nine billion thirty-seven thousand four. 

21. Eight hundred sixty-four billion five hundred thirty-eight 
million two hundred seventeen thousand nine hundred fifty-three. 



22 SIMPLE NUMBERS. 

22. One hundred seventeen quadrillion two hundred thirty-five 
trillion one hundred four billion seven hundred fifty million sixty- 
six thousand ten. 

23. Ninety-nine quintillion seven hundred forty-one trillion 
fifty-four billion one hundred eleven million one hundred one. 

24. One hundred octillion one hundred septillion one hundred 
quintillion one hundred quadrillion one hundred trillion one hundred 
billion one hundred million one hundred thousand one hundred. 

25. Four decillion seventy-five nonillion three octillion fifty- 
two septillion one sextillion four hundred seventeen quintillion 
ten quadrillion twelve trillion fourteen billion three hundred sixty 
million twenty-two thousand five hundred nineteen. 

Write the following numbers in figures, and read them : 

26. Twenty-five units in the 2d period, four hundred ninety-six 
in the 1st. Am. 25,496. 

27. Three hundred sixty-four units in the 3d period, seven 
hundred fifteen in the 2d, eight hundred thirty-two in the 1st. 

28. Four hundred thirty-six units in the 4th period, twelve in 
the 3d, one hundred in the 2d, three hundred one in the 1st. 

29. Eighty-one units in the 5th period, two hundred nineteen 
in the 4th, fifty-six in the 2d. 

30. Nine hundred forty-five units in the 7th period, eighteen in 
the 5th, one hundred three in the 3d. 

31. One unit in the 10th period, five hundred thirty-six in the 
9th, two hundred forty-seven in the 8th, nine hundred twenty-four 
in the 7th. 

Point off and read the following numbers : 



32. 


564. 


37. 


2005. 


33. 


24835. 


38. 


100103. 


34. 


2474783. 


39. 


53000008. 


35. 


247843112. 


40. 


1001005003. 


36. 


23678542789. 


41. 


750000040003. 




42. 247364582327896438542721. 






43. 379403270506038009503070. 




44. 


200057000320046730004 


30512500000567304705030040. 



ADDITION. 23 



ADDITION. 

©§g. Addition is the process of uniting several numbers of the 
same kind into one equivalent number. 

©3. The Sum or Amount is the result obtained by the process 
of addition. 

©4L When the given numbers contain several orders of units, 
the method of addition is based upon the following principles : 

I. If the like orders of units be added separately, the sum of 
all the results must be equal to the entire sum of the given num- 
bers. (Ax. 10). 

II. If the sum of the units of any order contain units of a 
higher order, these higher units must be combined with units of 
like order. Hence, 

III. The work must commence with the lowest unit, in order 
to combine the partial sums in a single expression, at one ope- 
ration. 

1. Find the sum of 397^476, and 873. 

operation Analysis. We arrange the numbers so that 

397 units of like order shall stand in the same column. 

476 We then add the first, or right hand column, and 

873 find the sum to be 16 units, or 1 ten and 6 units ; 

1746 writing the 6 units under the column of units, we 

add the 1 ten to the column of tens, and find the 

sum to be 24 tens, or 2 hundreds and 4 tens ; writing the 4 tens under 

the column of tens, we add the 2 hundreds to the column of hundreds, 

and find the sum to be 17 hundreds, or 1 thousand and 7 hundreds ; 

writing the 7 hundreds under the column of hundreds, and the 1 in 

thousands' place, we have the entire sum, 1746. 

©5. From these principles we deduce the following 

Kule. I. Write the numbers to be added so that all the units 

of the same order shall stand in the same column; that is, units 

under units , tens under tens, etc. 

II. Commencing at units, add each colum,n separately, and write 

the sum underneath, if it be less than ten. 



24 SIMPLE NUMBERS. 

III. If the sum of any column be ten or more than ten y write the 
unit figure only, and add the ten or tens to the next column. 

IV. Write the entire sum of the last column. 

Notes. i. In adding, learn to pronounce the partial results without naming 

the figures separately. Thus, in the operation given for illustration, say 3, 9, 
16; 8,15,24; 10,14,17. 

2. When the sum of any column is greater than 9, the process of adding the 
tens to the next column is called carrying. 

66. Proof. There are two principal methods of proving 
Addition. 

1st. By varying the combinations. 

Begin with the right hand or unit column, and add the figures 
in each column in an opposite direction from that in which they 
were first added ) if the two results agree, the work is supposed 
to be right. 

2d. By excess of 9's. 

67. This method depends upon the following properties of the 
number 9 : * 

L ' If a number be divided by 9, the remainder will be the same 
as when the sum of its digits is divided by 9. Therefore, 

II. If several numbers be added, the excess of 9's in the sum 
must be equal to the excess of 9's in the sum of all the digits in 
the numbers. 

1. Add 34852, 24784, and 72456, and prove the work by the 
excess of 9's. 

OPERATION. 

34852 
24784 
72456 ... 8, excess of 9's in all the digits of the numbers. 



132092 ...8, " " " " sum " " 

Analysis. Commencing with the first number, at the left hand, we 
say 3 and 4 are 7, and 8 are 15 ; dropping 9, the excess is 6, which 
added to 5, the next digit, makes 11 ; dropping 9, the excess is 2 ; 
then 2 and 2 are 4, and 2 (the left hand digit of the second number) 

* For a demonstration of these properties, see 186, IX. 



ADDITION. 25 

are 6, and 4 are 10 ; dropping 9, the excess is 1. Proceeding in like 
manner through all the digits, the final excess is 8 ; and as 8 is also 
the excess of 9 ? s in the sum, the work of addition is correct. It is 
evident that the same result will be obtained by adding the digits in 
columns as in rows. Hence, to prove Addition by excess of 9's : — 
Commencing at any figure, add. the digits of the given numbers 
in any order, dropping 9 as often as the amount exceeds 9. If 
the final excess be equal to the excess of 9's in the sum, the work 
is right. 

Note. — This method of proving addition by the excess of 9's, fails in the fol- 
lowing cases : 1st, when the figures of the answer are misplaced; 2d, when the 
value of one figure is as much too great as that of another is too small. 

EXAMPLES FOR PRACTICE. 



(1.) 


(2.) 


(3.) 


(4) 


8635 


1234567 


67 


24603 


2194 


723456 


123 


298765 


7421 


34565 


4567 


47321 


5063 


45666 


89093 


58653 


2196 


333 


654321 


5376 


1245 


90 


1234567 


340 



26754 2038677 

5. 123+ 456+785+12+ 345+901+567=how many? 

6. 12345+67890+8763+347+1037+198760=how many? 

7. 172+4005+3761+20472+367012+19762=how many? 

8. What is Ihe sum of thirty-seven thousand six, four hundred 
twenty-nine thousand nine, and two millions thirty-six ? 

Arts. 2,466,051. 

9. Add eight hundred fifty-six thousand nine hundred thirty- 
three, one million nine hundred seventy-six thousand eight hun- 
dred fifty-nine, two hundred three millions eight hundred ninety- 
five thousand seven hundred fifty-two. Am. 206,729,544. 

10. What is the sum of one hundred sixty-seven thousand, 
three hundred sixty-seven thousand, nine hundred six thousand, 
two hundred forty-seven thousand, seventeen thousand, one hun- 
dred six thousand three hundred, forty thousand forty-nine, ten 
thousand four hundred one ? Ans. 1,860,750. 

11. What number of square miles in New England, there 
3 



26 SIMPLE NUMBERS. 

being in Maine 31766, in New Hampshire 9280, in Vermont 
10212, in Massachusetts 7800, in Rhode Island 1306/ and in 
Connecticut 4674? Ans. 65,038. 

12. The estimated population of the above States, in 1855, was 
as follows; Maine 653000, New Hampshire 338000, Vermont 
327000, Massachusetts 1133123, Rhode Island 166500, and Con- 
necticut 384000. What was the entire population ? 

13. At the commencement of the year 1858 there were in ope- 
ration in the New England States, 3751 miles of railroad; in 
New York, 2590 miles; in Pennsylvania, 2546; in Ohio, 2946; 
in Virginia, 1233 ; in Illinois, 2678 ; and in Georgia, 1233. What 
was the aggregate number of miles in operation in all these States ? 

14. The Grand Trunk Railway is 962 miles long, and cost 
$60000000; the Great Western Railway is 229 miles long, and 
cost $14000000; the Ontario, Simcoe and Huron, is 95 miles 
long, and cost $3300000 ; the Toronto and Hamilton is 38 miles 
long, and cost $2000000. What is the aggregate length, and 
what the cost, of these four roads ? 

Ans. Length, 1,324 miles; cost, $79,300,000. 

15. A man bequeathed his estate as follows; to each of his 
two sons, $12450; to each of his three daughters, $6500; to 
his wife, $650 more than to both the sons, and the remainder, 
which was $1000 more than he had left to all his family, he gave 
to benevolent institutions. What was the whole amount of his 
property? Ans. $140,900. 

16. How many miles from the southern extremity of Lake 
Michigan to the Gulf of St. Lawrence, passing through Lake 
Michigan, 330 miles ; Lake Huron, 260 miles ; River St. Clair, 24 
miles; Lake St. Clair, 20 miles; Detroit River, 23 miles; Lake 
Erie, 260 miles; Niagara River, 34 miles; Lake Ontario, 180 
miles ; and the River St. Lawrence, 750 miles ? 

17. The United States exported molasses, in the year 1856, to 
the value of $154630; in 1857, $108003; in 1858, $115893; 
and tobacco, during the same years respectively, to the value of 
$1829207, $1458553, and $2410224. What was the entire value 
of the molasses and tobacco exported in these three years ? 



ADDITION. 27 

18. The population of Boston, in 1855, was 162629 ; Provi- 
dence, 50000; New York, 629810; Philadelphia, 408815; Brook- 
lyn, 127618 ; Cleveland, 43740 ; and New Haven, 25000. What 
was the entire population of these cities ? Ans. 1,447,612. 

19. Iron was discovered in Greece by the burning of Mount 
Ida, B. C. 1406 ; and the electro-magnetic telegraph was invented 
by Morse, A. D. 1832. What period of time elapsed between the 
two events ? Ans. 3,238 years. 

20. The number of pieces of silver coin made at the United 
States Mint at Philadelphia, in the year 1858, were as follows : 
4628000 half dollars, 10600000 quarter dollars, 690000 dimes, 
4000000 half dimes, and 1266000 three-cent pieces. What was 
the total number of pieces coined ? 

21. The cigars imported by the United States, in the year 1856, 
were valued at $3741460; in 1857, at $4221096; and in 1858, at 
$4123208. W T hat was the total value of the importations for the 
three years ? Ans. $12,085,764. 

22. In the appropriations made by Congress for the year ending 
June 30, 1860, were the following; for salary and mileage of 
members of Congress, $1557861 ; to officers and clerks of both 
Houses, $157639 ; for paper and printing of both Houses, $170000 ; 
to the President of the United States, $31450 ; and to the Vice 
President, $8000. What is the total of these items ? 



ADDING TWO OR MORE COLUMNS AT ONE OPERATION. 

68. 1. What is the sum of 4632, 2553, 4735, and 2863 ? 

operation. Analysis. Beginning with the units and tens of 

4632 the number last written, we add first the tens above, 

2553 then the units, thus ; 63 and 30 are 93, and 5 are 98, 

4735 and 50 are 148, and 3 are 151, and 30 are 181, and 

286 3 2 are 183. Of this sum, we write the 83 under the 

14783 columns added, and carry the 1 to the next columns, 

thus ; 28 and 1 are 29, and 40 are 69, and 7 are 76, 

and 20 are 96, and 5 are 101, and 40 are 141, and 6 are 147, which 

we write in its place, and we have the whole amount, 14783. 



28 



SIMPLE NUMBERS. 



EXAMPLES FOR PRACTICE. 



(1-) 


(2.) 


(3.) 


(4-) 


8450 


75634 


123456 


7349042 


5425 


86213 


47021 


2821986 


8595 


92045 


82176 


1621873 


6731 


73461 


570914 


236719 


7963 


34719 


379623 


401963 


5143 


26054 


7542 


67254 


4561 


19732 


25320 


45067 


6783 


84160 


57644 


910732 


4746 


97013 


908176 


6328419 


2373 


34567 


73409 


1437651 


3021 


43651 


3147 


9716420 


7273 


52170 


67039 


3191232 



71064 



719419 



2345467 



34128358 



5. What is the total number of churches, the number of persons 
accommodated, and the value of church property in the United 
States, as shown by the following statistics ? 



No. of 
churches. 

Methodist 12484 

Baptist 8798 

Presbyterian 4591 

Congregational 1675 

Episcopal 1430 

Roman Catholic 1269 

Lutheran 1205 

Christians 812 

Friends 715 

Union 619 

Universalist 494 

Free Church 361 

Moravian 331 

German Reformed 327 

Dutch Reformed 324 

Unitarian 244 

Mennonite 110 

Tunkers 52 

Jewish 31 

Swedenborgian 15 



No. of persons 


Value of 


iccommodated. 


church property. 


4220293 


$14636671 


3134438 


10931382 


2045516 


14469889 


795677 


7973962 


631613 


11261970 


705983 


8973838 


532100 


2867886 


296050 


845810 


283023 


1709867 


213552 


690065 


205462 


1767015 


108605 


252255 


112185 


443347 


156932 


965880 


181986 


4096730 


137867 


3268122 


29900 


94245 


35075 


46025 


16575 


371600 


5070 


108100 



ADDITION. 29 

6. Give the amounts of the productions of the United States and 
Territories for the year 1850, as expressed in the following columns : 

Pounds of 
Butter. 

Alabama 4,008,811 

Arkansas 1,854,239 

California 705 

Columb.,Dist. 14,872 
Connecticut... 6,498,119 

Delaware 1,055,308 

Florida 371,498 

Georgia 4,640,559 

Illinois 12,526,543 

Indiana 12,881,535 

Iowa 2,171,188 

Kentucky.... 9,947,523 

Louisiana 683,069 

Maine 9,243,811 

Maryland 3,086,160 

Massachusetts 8,071,370 

Michigan 7,065,878 

Mississippi... 4,346,234 

Missouri 7,834,359 

N.Hampshire 6,977,056 
New Jersey... 9,487,210 
New York.... 79,766,094 
N.Carolina... 4,146,290 

Ohio 34,449,379 

Pennsylvania 39,878,418 
Rhode Island 995,670 
S. Carolina... 2,981,850 
Tennessee.... 8,139,585 

Texas 2,344,900 

Vermont 12,137,980 

Virginia 11,089,359 

Wisconsin.... 3,633,750 
Territories... 295,984 



3* 



O 



Pounds of 
Cheese. 


Founds of 
Wool. 


Bushels of 
Wheat. 


31,412 


657,118 


294,044 


30,088 


182,595 


199,639 


150 


5,520 


17,228 


1,500 


525 


17,370 


5,363,277 


497,454 


41,762 


3,187 


57,768 


482,511 


18,015 


23,247 


1,027 


46,976 


990,019 


1,088,534 


1,278,225 


2,150,113 


9,414,575 


624,564 


2,610,287 


6,214,458 


209,840 


373,898 


1,530,581 


213,954 


2,297,433 


2,142,822 


1,957 


109,897 


417 


2,434,454 


1,364,034 


296,259 


3,975 


477,438 


4,494,680 


7,088,142 


585,138 


31,211 


1,011,492 


2,043,283 


4,925,889 


21,191 


559,619 


137,990 


203,572 


1,627,164 


2,981,652 


3,196,563 


1,108,476 


185,658 


365,756 


375,396 


1,601,190 


t9,741,413 


10,071,301 


13,121,498 


95,921 


970,738 


2,130,102 


10,819,542 


10,196,371 


14,487,351 


2,505,034 


4,481,570 


15,367,691 


316,508 


129,692 


49 


4,970 


487,233 


1,066,277 


177,681 


1,364,378 


1,619,386 


95,299 


131,917 


41,729 


8,720,834 


3,400,717 


535,955 


436,292 


2,860,765 


11,212,616 


400,283 


253,963 


4,286,131 


73,826 


71,894 


517,562 



30 SIMPLE NUMBERS. 



SUBTKACTIOK 

69. Subtraction is the process of taking one number from 
another of the same kind, equal to or greater than itself. 

70. The Minuend is the number to be subtracted from. 

71. The Subtrahend is the number to be subtracted. 

73. The Difference or Remainder is the result obtained by 
the process of subtraction. 

73. When the given numbers contain more than one figure 
each, the method of subtraction depends upon the following prin- 
ciples : 

I. If the units of each order in the subtrahend be taken sepa- 
rately from the units of like order in the minuend, the sum of the 
differences must be equal to the entire difference of the given 
numbers. (Ax. 11.) 

II. If both minuend and subtrahend be equally increased, the 
remainder will not be changed. 

1. From 928 tale 275. 

operation. Analysis. We first subtract 5 units from 

Minuend, 928 8 units, and obtain 3 units for a partial re- 

Subtrahend, 275 mainder. As we cannot take 7 tens from 2 

Remainder 653 tens, we add 10 tens to the 2 tens, making 
12 tens ; then 7 tens from 12 tens leave 5 
tens, the second partial remainder. Now, since we added 10 tens, 
or 1 hundred, to the minuend, we will add 1 hundred to the subtra- 
hend, and the true remainder will not be changed (II) ; thus, 1 
hundred added to 2 hundreds makes 3 hundreds, and this sum sub- 
tracted from 9 hundreds leaves 6 hundreds ; and we have for the total 
remainder, 653. 

Note. — The process of adding 10 to the minuend is sometimes called bor- 
rowing 10, and that of adding 1 to the next figure of the subtrahend, carrying 1. 

74. From these principles and illustrations we deduce the 
following 

Hule. I. Write the less number under the greater, placing units 
of the same order under each other. 



SUBTRACTION. 31 

II. Begin at the right hand, and take each figure of the subtra- 
hend from the figure above it, and write the result underneath. 

III. If any figure in the subtrahend be greater than the corres- 
ponding figure above it, add 10 to that upper figure before sub- 
tr acting, and then add 1 to the next left hand figure of the subtra- 
hend. 

75, Proof. It is evident that the subtrahend and remainder 
must together contain as many units as the minuend ; hence, to 
prove subtraction, we have three methods : 

1st. Add the remainder to the subtrahend; the sum will be 
equal to the minuend. Or, 

2d. Subtract the remainder from the minuend ; the difference 
will be equal to the subtrahend. Or, 

3d. Find the excess of 9's in the remainder and subtrahend 
together, and it will be equal to the excess of 9's in the minuend. 

EXAMPLES FOR PRACTICE. 

(1.) (2.) (3.) (4.) 
From 47965 103767 57610218 89764321 
Take - 26714 98731 8306429 83720595 

Rem. 21251 5036 49303789 6043726 

5. From 180037561 take 5703746. 

6. From 2460371219 take 98720342. 

7. 89037426175 — 2435036749 = how many? 

8. 10000033421 — 999044110 == how many? 

9. A certain city contains 146758 inhabitants, which is 3976 
more than it contained last year; how many did it contain last 
year? Ans. 142,782. 

10. The first newspaper published in America was issued at 
Boston in 1704; how long was that before the death of Benjamin 
Franklin, which occurred in 1790 ? 

11. A merchant sold a quantity of goods for $42017, which 
was $1675 more than they cost him; how much did they cost 
him? Ans. $40,342. 

12. In 1858 the exports of the United States amounted to 



32 SIMPLE NUMBERS. 

$324644421, and the imports to $282613150; how much did the 
exports exceed the imports ? Ans. $42,031,271. 

13. In 1858 the gold coinage of the United States amounted to 
$52889800, and the silver to $8233287; how much did the gold 
exceed the silver coinage ? 

14. The South in 1850 produced 978311690 pounds of cotton, 
valued at $101834616, and 237133000 pounds of sugar valued at 
$16599310; how much did the cotton exceed the sugar in quan- 
tity and in value? Ans. 741,178,690 pounds; $85,235,306. 

15. The area of the Chinese Empire is 5110000 square miles, 
and that of the United States 2988892 square miles ; the esti- 
mated population of the former is 340000000, and that of the 
latter in 1850 was 23363714. What is the difference in area and 
in population? 

16. The population of London in 1850 was 2362000, and that 
of New York city 515547 ; how many more inhabitants had London 
than New York ? Ans. 1,846,453. 

17. The total length of railroads in operation in the United 
States, January 1, 1859, was 27857 miles, and the total length of 
the canals 5131 miles ; how many miles more of railxoad than of 
canal? Ans. 22,726. 

18. The entire deposit of domestic gold at the United States 
Mint and its branches, to June, 1859, was $470341478, of which 
$451310840 was from California; how much was received from 
other sources ? Ans. $19,030,638. 

19. During the year ending September 30, 1858, the number 
of letters exchanged between the United States and Great Britain 
were 1765015 received, and 1603609 sent; between the United 
States and France, 624795 received, and 639906 sent. How many 
letters did the exchange with Great Britain exceed those with 
France? Ans. 2,103,923. 

20. The Southern States in 1850 had a population of 6696061, 
the Middle States 6624988, and the Eastern States 2728116; 
how many more inhabitants had the Middle and Eastern States 
than the Southern States ? 

21. Having $20000, I wish to know how much more I must 



SUBTRACTION. 



33 



accumulate to be able to purchase a piece of property worth $23470, 
and have $5400 left? Ans. $8,870. 

22. A has $3540 more than B, and $1200 less than C, who has 
$20600 ; D has as much as A and B together. How much has D ? 

Ans. $35,260. 

TWO OR MORE SUBTRAHENDS. 

76. Two or more numbers may be taken from another at a 
single operation, as shown by the following example : 

1. A man having 1278 barrels of flour, sold 236 barrels to A, 
362 to B, and 387 to C \ how many had he left ? 



OPERATION. 



Minuend, 



Subtrahends, 



Remainder, 



1278 




Analysis. Since the remainder sought, 
added to the subtrahends, must be equal to 
the minuend, we add the columns of the 
subtrahends, and supply such figures in the 
remainder as, combined with these sums, 
will produce the minuend. Thus, 7 and 2 
are 9, and 6 are 15, and 3 (supplied in the 
remainder sought) are 18 ; then, carrying 
the tens' figure of the 18, J. and 8 are 9, and 6 are 15, and 3 are 18, 
and 9 (supplied in the remainder) are 27; lastly, 2 to carry to 3 are 
5, and 3 are 8, and 2 are 10, and 2 (supplied in the remainder) are 
12; and the whole remainder is 293. Hence, the following 

Rule. T. Having written the several subtrahends under the min- 
uend, add the first column of the subtrahends, and supply such a 
figure in the remainder sought, as, added to this partial sum, will 
give an amount having for its unit figure the figure above in the 
minuend. 

II. Carry the tens 1 figure of this amount to the next column of 
the subtrahends, and proceed as before till the entire remainder is 
obtained. 

EXAMPLES FOR PRACTICE. 



From 


a-) 

47962 


(2-) 
127368 


(3.) 
903486 


(4-) 
2503734 


Take \ 


21435 

15672 

456 


56304 
4782 
9156 


430164 

132875 

67321 


89763 

94207 

237561 


Rem. 


10399 


57126 


273126 


2082200 



34 SIMPLE NUMBERS. 

5. From 4568 take 1320 + 275 + 320. 

6. Subtract 1200 + 750 + 96 from 4756 + 575 + 140 + 84. 

7. A man bought four city lots, for which he paid $15760. For 
the first he paid $2175, for the second $3794, and for the third 
$4587; how much did he pay for the fourth ? Ans. $5,204. 

8. John Wise owns property to the amount of $75860, of which 
he has $45640 invested in real estate, $25175 in personal property, 
and the remainder he has in bank ; how much has he in bank ? 

9. Lake Huron contains 20000 square miles; by how much 
does it exceed the area of Lake Erie and Lake Ontario, the former 
containing 11000 square miles, and the latter 7000 ? 

Ans. 2000 square miles. 

10. In the year 1852, there arrived in the United States 398470 
immigrants, of whom 157548 were born in Ireland, and 143429 
were born in Germany; how many were born in other countries? 

Ans. 97,493. 

11. The entire amount of coinage in the United States for the 
year ending June, 1858, was $61357088, of which $52889800 was 
of gold, $234000 of copper, and the remainder of silver; how much 
was of silver ? 

12 A speculator gained $5760, and afterward lost $2746 ; at 
another time he gained $3575, and then lost $4632. How much 
did his gains exceed his losses ? Ans. $1,957. 

13. The Eastern States have an area of 65038 square miles, 
the Middle States 114624 square miles, and the Southern States 
643166 square miles; how many more square miles have the 
Southern than the Middle and Eastern States ? 

14. The entire revenue of the United States Post Office Depart- 
ment for the year ending Sept. 30, 1858, was $8186793, of which 
sum $5700314 was received for stamps and stamped letters, and 
$904299 for letter-postage in money; how much was received from 
all other sources ? Ans. $1,582,180. 

15. The total expenditures ot the Department for the same year 
were $12722470, of which sum $7821556 was paid for the trans- 
portation of inland mails, $424497 for the transportation of foreign 
mails and $2355016 as compensation to postmasters; how much 
was expended for all other purposes? $2,121,401. 



MULTIPLICATION. 



35 



MULTIPLICATION. 

77. Multiplication is the process of taking one of two given 
numbers as many times as there are units in the other. 

78. The Multiplicand is the number to be taken. 

79. The Multiplier is the number which shows how many 
times the multiplicand is to be taken. 

SO, The Product is the result obtained by the process of mul- 
tiplication. 

81. The Factors are the multiplicand and multiplier. 

Notes. — 1. Factors are producers, and the multiplicand and multiplier are 
called factors because they produce the product. 

2. Multiplication is a short method of performing addition when the numbers 
to be added are equal. 

82. The method of multiplying when either factor contains 
more than one figure, depends upon the following principles : 

It is evident that 5 units taken 3 times is the same as 3 units 
taken 5 times ; and the same is true of any two factors. Hence, 

I. The product of any two factors is the same, whichever is used 
as the multiplier. If units be multiplied by units, the product 
will be units ; if tens be multiplied by units, or units by tens, the 
product will be tens ; and so on. That is, 

II. If either factor be units of the first order, the product will 
be units of the same order as the other factor. 

III. If the units of each order in the multiplicand be taken sepa- 
rately as many times as there are units in the multiplier, the sum 
of the products must be equal to the entire product of the given 
numbers (Ax. 10). 

1. Multiply 346 by 8. 

operation. Analysis. In this example it is re- 

Muitipiicand, 346 quired to take 346 eight times. If we take 

Multiplier, 8 the units of each order 8 times, we shall 

Product 97P8 ta ^ e ^ e en ^ re num ber 8 times, (III). 

Therefore, commencing at the right hand, 

we say ; 8 times 6 units are 48 units, or 4 tens and 8 units ; writing 

the 8 units in the product in units' place, we reserve the 4 tens to 

add to the next product ; 8 times 4 tens are 32 tens, and the 4 tens 



36 



SIMPLE NUMBERS. 



reserved in the last product added, are 36 tens, or 3 hundreds and 6 
tens ; we write the 6 tens in the product in tens' place, and reserve 
the 3 hundreds to add to the next product ; 8 times 3 hundreds are 
24 hundreds, and the 3 hundreds reserved in the last product added, 
are 27 hundreds, which being written in the product, each figure in 
the place of its order, gives, for the entire product, 2768. 

2. Multiply 758 by 346. 

operation. Analysis. In this example the multiplicand is 

75Q to he taken 346 times, which may be done by 

346 taking the multiplicand separately as many times 

777o as there are units expressed by each figure of the 

oaoo multiplier. 758 multiplied by 6 units is 4548 

2274 units, (II) ; 758 multiplied by 4 tens is 3032 tens, 

9A99fi£ (^)» wn * cn we write w ^h its lowest order in tens' 

place, or under the figure used as a multiplier ; 

758 multiplied by 3 hundreds is 2274 hundreds, (II), which we write 

with its lowest order in hundreds' place. Since the sum of these 

products must be the entire product of the given numbers, (III), we 

add the results, and obtain 262268, the answer. 

Notes. — 1. When the multiplier contains two or more figures, the several re- 
sults obtained by multiplying by each figure are called partial products. 

2. When there are ciphers between the significant figures of the multiplier, 
Dass over them, and multiply by the significant figures only. 

83. From these principles and illustrations we deduce the fol- 
lowing general 

Rule. I. Write the multiplier under the multiplicand, placing 
units of the same order under each other. 

II. Multiply the multiplicand by each figure of the multiplier 
successively , beginning with the unit figure , and write the first figure 
of each partial product under the figure of the multiplier used, 
writing down and carrying as in addition. 

III. If there are partial products, add them, and their sum will 
be the product required. 

Note. — The multiplier denotes simply the number of times the multiplicand 
is to be taken ; hence, in the analysis of a problem, the multiplier must be con- 
sidered as abstract, though the multiplicand may be either abstract or concrete. 

84. Proof. There are two principal methods of proving 
multiplication. 



MULTIPLICATION. 37 

1st. By varying the partial products. 

Invert the order of the factors ; that is, multiply the multiplier 
by the multiplicand : if the product is the same as the first result, 
the work is correct. 

2d. By excess of 9's. 

85. The illustration of this method depends upon the following 
principles : 

I. If the excess of 9's be subtracted from a number, the re- 
mainder will be a number having no excess of 9's. 

II. If a number having no excess of 9's be multiplied by any 
number, the product will have no excess of 9 J s. 

1. Let it be required to multiply 473 by 138. 

Analysis. The excess of 
9's in 473 is 5, and 473 = 468 
+ 5, of which the first part, 
468, contains no excess of 9's, 
(I). The excess of 9's in 
138 is 3, and 138 = 135 + 3, of 
which the first part, 135, con- 
tains no excess of 9's, (I). 
Multiplying both parts of the 
multiplicand by each part of the multiplier, we have four partial pro- 
ducts, of which the first three have no excess of 9 ? s, because each con- 
tains a factor having no excess of 9's, (II). Therefore, the excess 
of 9's in the entire product must be the same as the excess of 9's in 
the last partial product, 15, which we find to be 1 + 5 = 6. The same 
may be shown of any two numbers. Hence, to prove multiplication 
by excess of 9's, 

Find the excess of 9's in each of the two factors, and multiply 
them together; if the excess of 9's in this product is equal to the 
excess of 9's in the product of the factors, the work is supposed 
to be right. 

Note. — If the excess of 9's in either factor is 0, the excess of 9's in the pro- 
duct will be 0, (II). 

EXAMPLES FOR PRACTICE. 





OPERATION. 


473 = 468 + 5 
138 = 135 + 3 


Partial 
products. 


f 468 x 135 = 63180 

5 x 135 = 675 

468 x 3 = 1404 

5x3= 15 


Entire prodi 


ict, '65274 



(10 


(2.) 


(3.) 


(40 


475 


3172 


9827 


7198 


9 


14 


84 


216 



Multiply 

By 

Prod. 4275 44408 825468 1554768 

4 



38 SIMPLE NUMBERS. 

5. Multiply 31416 by 175. Ans. 5497800, 

6. Multiply 40930 by 779. Ans. 31884470. 

7. Multiply 46481 by 936. 

8. Multiply 15607 by 3094. 

9. Multiply 281216 by 978. Ans. 27502924o 

10. Multiply 30204 by 4267. Ans. 128,880,468. 

11. What is the product of 4444 x 2341 ? 

Ans. 10,403,404. 

12. What is the product of 4567 x 9009 ? 

Ans. 41,144,103. 

13. What is the product of 2778588 x 9867? 

Ans. 27,416,327,796. 

14. What is the product of 7060504 x 30204 ? 

Ans. 213,255,462,816. 

15. What will be the cost of building 276 miles of railroad at 
$61320 per mile ? Ans. $16,924,320. 

16. If it require 125 tons of iron rail for one mile of railroad, 
how many tons will be required for 196 miles ? 

17. A merchant tailor bought 36 pieces of broadcloth, each 
piece containing 47 yards, at 7 dollars a yard ; how much did he 
pay for the whole ? Ans. $11,844. 

18. The railroads in the State of New York, in operation in 
1858, amounted to 2590 miles in length, and their average cost 
was about $52916 per mile ; what was the total cost of the rail- 
roads in New York ? Ans. $137,052,440. 

19. The Illinois Central Railroad is 700 miles long, and cost 
$45210 per mile ; what was its total cost ? 

20. The salary of a member of Congress is $3000, and in 1860 
there were 303 members ; how much did they all receive ? 

21. The United States contain an area of 2988892 square miles, 
and in 1850 they contained 8 inhabitants to each square mile; 
what was their entire population ? Ans. 23,911,136. 

22. Great Britain and Ireland have an area of 118949 square 
miles, and in 1850 they contained a population of 232 to the square 
mile; what was their entire population? Ans. 27,596,168. 

23. The national debt of France amounts to $32 for each indi- 



MULTIPLICATION. 39 

yidual, and the population in 1850 was 35781628 ; what was the 
entire debt of France ? Ans. 1,145,012,096. 

POWERS OP NUMBERS. 

86. We have learned (15) that a power is the product arising 
from multiplying a number by itself, or repeating it any number 
of times as a factor ; (1G), that a root is a factor repeated to pro- 
duce a power; and (40) an index or exponent is the number in- 
dicating the power to which a number is to be raised. 

87. The First Power of any number is the number itself, or 
the root; thus, 2, 3, 5, are first powers or roots. 

88. The Second Power, or Square, of a number is the pro- 
duct arising from using the number two times as a factor; thus, 
2 2 ^2x2 = 4; 5 2 =5x5 = 25. 

89. The Third Power, or Cube, of a number is the product 
arising from using the number three times as a factor; thus, 
4 3 =4 x 4 x 4 = 64. 

90. The higher powers are named in the order of their num- 
bers, as Fourth Power, Fifth Power, Sixth Potver, etc. 

91. 1. What is the third power or cube of 23 ? 

operation. Analysis. We multiply 23 

23 x 23 x 23 = 12167 D J 23 > and the P r °duct by 23; 

and, since 23 has been taken 3 
times as a factor, the last product, 12167, must be the third power or 
cube of 23. Hence, 

Rule. Multiply the number by itself as many times, less 1, as 
there are units in the exponent of the required power. 

Note. — The process of producing any required power of a number by multi- 
plication is called Involution* 

EXAMPLES FOR PRACTICE. 

1. What is the square of 72 ? Ans. 5184. 

2. What is the fifth power of 12 ? Ans. 248832. 

3. What is the cube of 25 ? 

4. What is the seventh power of 7 ? Ans. 823543. 



40 SIMPLE NUMBERS. 

5. What is the fourth power of 19 ? Ans. 130321. 

6. Required the sixth power of 3. Ans. 729. 

7. Find the powers indicated in the following expressions : 
9 5 , IP, 18 2 , 125 4 , 786 2 , 94 6 , 100*, 17 3 , 251. 5 

8. Multiply 8 3 by 15 2 . Ans. 115200. 

9. What is the product of 25 2 X 3 4 ? 

10. 7 3 X 200 = 4 4 x ll 2 , and how many ? Ans. 37,624. 

GENERAL PRINCIPLES OF MULTIPICATION. 

02. There are certain general principles of multiplication, of 
use in various contractions and applications which occur in sub- 
sequent portions of this work. These relate, 1st, to changing the 
factors by addition or subtraction ; 2d, to the use of successive 
factors in continued multiplication. 

CHANGING THE FACTORS BY ADDITION OR SUBTRACTION. 

03. The product is equal to either factor taken as many times 
as there are units in the other factor. (82, I). Hence, 

I. Adding 1 to either factor, adds the other factor to the pro- 
duct. 

11. Subtracting 1 from either factor, subtracts the other factor 
from the product. Hence, 

III. Adding any number to either factor, increases the pro- 
duct by as many times the other factor as there are units in the 
number added; and subtracting any number from either factor, 
diminishes the product by as many times the other factor as there 
are units in the number subtracted. 

CONTINUED MULTIPLICATION. 

04. A Continued Multiplication is the process of finding the 
product of three or more factors, by multiplying the first by the 
second, this result by the third, and so on. 

05. To show the nature of continued multiplication, we 
observe : 

1st. If any number, as 17, be multiplied by any other number, 
as 3, the result will be 3 times 17; if this result be multiplied by 



MULTIPLICATION. 41 

another number, as 5, the new product will be 5 times 3 times 17, 
which is evidently 15 times 17. Hence, 17 X 3 x 5 = 17 X 15; 
the same reasoning would extend to three or more multipliers. 

2d. Since 5 times 3 is equal to 3 times 5, (82, 1), it follows that 
17 multiplied by 5 times 3 is the same as 17 multiplied by 3 times 
5 ; or 17 X 3 x 5 = 17 X 5 x 3. Hence, the product is not 
changed by changing the orders of the factors. 

These principles may be stated as follows : 

I. If a given number be multiplied by several factors in con- 
tinued multiplication, the result will be the same as if the given 
number were multiplied by the product of the several multipliers. 

II. The product of several factors in continued multiplication 
will be the same, in whatever order the factors are taken. 

CONTRACTIONS IN MULTIPLICATION. 
CASE I. 

96. "When the multiplier is a composite number. 

A Composite Number is one that may be produced by multi- 
plying together two or more numbers. Thus, 18 is a composite 
number, since 6 x 3 = 18 ; or, 9 x 2 == 18 ; or, 3 x 3 X 2 == 18. 

97. The Component Factors of a number are the several 
numbers which, multiplied together, produce the given number; 
thus, the component factors of 20 are 10 and 2 (10 X 2 = 20); 
or, 4 and 5 (4 x 5 = 20); or, 2 and 2 and 5 (2 x 2 x 5 = 20). 

Note. — The pupil must not confound the factors with the parts of a number. 
Thus, the factors of which 12 is composed, are 4 and 3 (4 X 3 = 12) ; while the 
parts of which 12 is composed are 8 and 4 (8 + 4= 12) ; or 10 and 2(10 + 2 = 12). 
The factors are multiplied, while the parts are added, to produce the number. 

98. 1. Multiply 327 by 35. 

OPERATION. 

327 Analysis. The factors of 35 are 7 and 5. 

7 We multiply 327 by 7, and this result by 5, 

22gQ an d obtain 11445, which must be the same 

5 as the product of 327 by 5 times 7, or 35. 

(95, I). Hence we have the following 



11445 
4* 



42 SIMPLE NUMBERS. 

Rule. I. Separate the composite number into two or more 
/actors. 

II. Multiply the multiplicand by one of these factors, and that 
product by another, and so on until all the factors have been used 
successively ; the last product will be the product required. 

Note. — The factors may be used in any order that is most convenient, (95, II). 
EXAMPLES FOR PRACTICE. 

1. Multiply 736 by 24. Ans. 17664. 

2. Multiply 538 by 56. Ans. 30128. 

3. Multiply 27865 by 84. 

4. Multiply 7856 by 144. Ans. 1131264. 

5. What will 56 horses cost at 185 each ? 

6. If a river discharge 17740872 cubic feet of water in one 
hour, how much will it discharge in 96 hours ? 

Ans. 1703123712 cubic feet. 

CASE II. 

99. "When the multiplier is a unit of any order. 

If we annex a cipher to the multiplicand, each figure is removed 
one place toward the left, and consequently the value of the whole 
number is increased tenfold, ((57), III). If two ciphers are 
annexed, each figure is removed two places toward the left, and 
the value of the number is increased one hundred fold ; and every 
additional cipher increases the value tenfold. Hence, the 

Rule. Annex as many ciphers to the multiplicand as there are 
ciphers in the multiplier. 

EXAMPLES FOR PRACTICE. 

1. Multiply 364 by 100. Ans. 36400. 

2. Multiply 248 by 1000. Ans. 248000. 

3. What cost 1000 head of cattle at 50 dollars each ? 

4. Multiply one million by one hundred thousand ? 

5. How many letters will there be on 100 sheets, if each sheet 
have 100 lines, and each line 100 letters ? Ans. 1000000. 



MULTIPLICATION. 43 

CASE III. 

100, When there are ciphers at the right hand of one 
or both of the factors. 

1. Multiply 7200 by 40. 

operation. Analysis. The multiplicand, factored, is 

7200 equal to 72 X 100; the multiplier, factored, is 

40 equal to 4 X 10 ; and as these factors taken in 

288000 ali y or( ^ er w iU gi ye tne same product, (95, II), 

we first multiply 72 by 4, then this product 
by 100 by annexing two ciphers, and this product by 10 by annexing 
one cipher. Hence, the following 

Rule. Multiply the significant figures of the multiplicand by 
those of the multiplier, and to the product annex as many ciphers 
as there are ciphers on the right of both factors. 

EXAMPLES FOR PRACTICE. 

1. Multiply 740 by 300. Ans. 222000. 

2. Multiply 36000 by 240. Ans. 8640000. 

3. Multiply 20700 by 500. 

4. Multiply 4007000 by 3002. Ans. 12029014000. 

5. Multiply 300200 by 640. 

CASE IV. 

101. "When one part of the multiplier is a factor of 
another part. 

1. Multiply 4739 by 357. 

operation. Analysis. In this example, 7, one 

a>jqq part of the multiplier, is a factor of 

0C7 35, the other part. We first find, in 

the usual manner, the product of the 

33173 Prod. by 7 units. mu i t ipli ca nd by the 7 units; niulti- 
165865 Prod ' by 35 tens ' plying this product by 5, and writing 
1691823 Ans. the first figure of the result in tens' 

place, we obtain the product of the 
multiplicand by 7 X 5 X 10 = 35 tens ; and the sum of these two par- 
tial products must be the whole product required. 



44 SIMPLE NUMBERS. 

2. Multiply 58327 by 21318. 

operation. Analysis. In this exam- 

r nnnn pie, the 3 hundreds is a factor 

2i o i o of 18, the part on the right of 

it, and also of 21, the part on 



174981 Prod, by 3 hundreds. the left of it. We first mul 

1049886 Prod, by 18 units. ^ ^ 

* 22486 7 ^od. o,2I thousand, ^ ^ ^^ ^ . 

1243414986 -4tis. multiplying this product by 

6, and writing the first figure 
in units' place, we obtain the product of the multiplicand by 3 X 6 = 
18 units ; multiplying the first partial product by 7, and writing the 
first figure in thousands' place, we obtain the product of the multipli- 
cand by 7 X 3 X 1000 = 21 thousands ; and the sum of these three 
partial products must be the entire product required. 

Note. — The product obtained by multiplying any partial product is called a 
derived product, 

102. From these illustrations we have the following 

Rule. I. Find the product of the multiplicand by some figure 

of the multiplier which is a factor of one or more parts of the 

multiplier. 

II. Multiply this product by that factor which, taken with the 
figure of the multiplier first used, will produce other parts of the 
multiplier, and write the first figure of each result under the first 
figure of the part of the multiplier thus used. 

III. In like manner, find the product, either direct or derived, 
for every figure or part of the multiplier; the sum of all the pro- 
ducts will be the whole product required. 

EXAMPLES FOR PRACTICE. 

1. Multiply 5784 by 246. Ans. 1422864. 

2. Multiply 3785 by 721. Ans. 2728985. 

3. Multiply 472856 by 54918. Ans. 25568305808. 

4. Multiply 43785 by 7153. Ans. 313194105. 

5. Multiply 573042 by 24816. Ans. 14220610272. 

6. Multiply 78563721 by 127369. 

7. Multiply 43725652 by 5187914. 



MULTIPLICATION. 45 

8. Multiply 3578426785 by 64532164. 

9. Multiply 2703605 by 4249784. 

10. What is the product of 9462108 multiplied by 16824? 

Arts. 159,190,504,992. 

EXAMPLES COMBINING THE PRECEDING RULES. 

1. A man bought two farms, one containing 175 acres at $28 
per acre, and the other containing 320 acres at $37 per acre; 
what was the cost of both ? Ans. $16,740. 

2. If a man receive $1200 salary, and pay $364 for board, 
$275 for clothing, $150 for books, and $187 for other expenses, 
how much can he save in 5 years ? Ans. $1,120. 

3. Two persons start from the same point, and travel in oppo- 
site directions; one travels 29 miles a day, and the other 32 miles. 
How far apart will they be in 17 days? Ans. 1,037 miles. 

4. A drover bought 127 head of cattle at $34 a head, and 97 
head at $47 a head, and sold the whole lot at $40 a head; what 
was his entire profit or loss ? Ans. $83 profit. 

5. Multiply 675 — (77 + 56) by (3 x 156) — (214 — 28). 

Ans. 152844. 

6. Multiply 98 + 6x (37 + 50) by (64 — 50) x 5 — 10. 

Ans. 37200. 

7. What is the product of (14 x 25) — (9"x¥) + 4324 x 



(280 — 112) + (376 + 42) x 4 ? Ans. 8,004,000. 

8. In 1850 South Carolina cultivated 29967 farms and planta- 
tions, containing an average of 541 acres each, at an average value 
of $2751 for each farm ; New Jersey cultivated 23905 farms, con- 
taining an average of 115 acres each, at an average value of 
$5030 per farm. How much more were the farming lands of the 
latter valued at, than those of the former ? 

9. There are in the United States 1922890880 acres of land; 
of this there were reported under cultivation, in 1850, 1449075 
farms, each embracing an average of 203 acres. How many acres 
were still uncultivated ? 

10. Each of the above farms in the United States were valued 
at an average of $2258, and upon each farm there was an average 



46 SIMPLE NUMBERS. 

of $105 in implements and machinery. What was the aggregate 
value of the farms and implements? Ans. $3,424,164,225. 

Find the values of the following expressions : 

11. (4 3 -v-2 2 ) X 5 5 — -7 3 ? Ans. 49,657. 

12. 15 3 — (3 2 x 2 5 ) + 208 2 — (18 4 -v- 9 3 ) ? Ans. 46,207. 

13. 2 2 + 3 3 + 4 4 + 5 5 -f 6 6 ? 

14. In 1852 Great Britain consumed 1200000 bales of American 
cotton ; allowing each bale to contain 400 pounds, what was its 
total weight? 

15. If a house is worth $2450, and the farm on which it stands 
6 times as much, lacking $500, and the stock on the farm twice 
as much as the house, what is the value of the whole ? 

Ans. 21550. 

16. A flour merchant bought 1500 barrels of flour at 7 dollars 
a barrel; he sold 800 barrels at 10 dollars a barrel, and the re- 
mainder at 6 dollars a barrel. How much was his gain ? 

17. A man invests in trade at one time $450, at another $780, 
at another $1250, and at another $2275 ; how much must he add 
to these sums, that the amount invested by him shall be increased 
fourfold? Ans. $14,265. 

18. At the commencement of the year 1858 there were in ope- 
ration in the United States 35000 miles of telegraph; allowing 
the average cost to be $115 per mile, what was the total cost? 

19. The cost of the Atlantic Telegraph Cable, as originally 
made, was as follows ; 2500 miles at $485 per mile, 10 miles deep- 
sea cable at $1450 per mile, and 25 miles shore ends at $1250 per 
mile. What was its total cost ? Ans. $1,258,250. 

20. For the year ending June 30, 1859, there were coined in 
the United States 1401944 double eagles valued at twenty dollars 
each, 62990 eagles, 154555 half eagles, and 22059 three dollar 
pieces; what was the total value of this gold coin? 

Ans. $29,507,732. 



division. 47 



DIVISION. 

103. Division is the process of finding how many times one 
number is contained in another. 

104. The Dividend is the number to be divided. , 
lOo. The Divisor is the number to divide by. 

106. The Quotient is the result obtained by the process of 
division, and shows how many times the divisor is contained in 
the dividend. 

107. The Reciprocal of a number is 1 divided by the number. 
Thus, the reciprocal of 15 is 1 -f- 15, or y 1 ^. 

Notes. — 1. When the dividend does not contain the divisor an exact number 
of times, the part of the dividend left is called the Remainder, which must be 
less than the divisor. 

2. As the remainder is always a part of the dividend, it is always of the same 
name or kind. 

3. When there is no remainder the division is said to be exact. 

108. The method of dividing any number by another depends 
upon the following principles : 

I. Division is the reverse of multiplication, the dividend cor- 
responding to the product, and the divisor and quotient to the 
factors. 

II. If all the parts of a number be divided, the entire number 
will be divided. 

Since the remainder in dividing any part of the dividend must 
be less than the divisor, it can be divided only by being expressed 
in units of a lower order. Hence, 

III. The operation must commence with the units of the high- 
Got order. 

. 1. Divide 2742 by 6. 

Analysis. We write the divisor at the left 
operation. . „ , ,. . , _ . ., _ ,. 

of the dividend, separated from it by a line. 

' As 6 is not contained in 2 thousands, we take 

457 Arts. the 2 thousands and 7 hundreds together, and 

proceed thus ; 6 is contained in 27 hundreds 

4 hundred times, and the remainder is 3 hundreds ; we write 4 in 

hundreds' place in the quotient, and unite the remainder, 3 hundreds, 



48 SIMPLE NUMBERS. 

to the next figure of the dividend, making 34 tens ; then, 6 is con- 
tained in 34 tens 5 tens times, and the remainder is 4 tens ; writing 
5 tens in its place in the quotient, we unite the remainder to the next 
figure in the dividend, making 42 ; 6 is contained in 42 units 7 times, 
and there is no remainder ; writing 7 in its place in the quotient, we 
have the entire quotient, 457. 

Note 1. — The different numbers which we divide in obtaining the successive 
figures of the quotient, are called partial dividends. 



2. Divide 18149 by 56. 

OPERATION. 



Analysis. As neither 1 nor 18 
will contain the divisor, we take 
56 ) 18149 ( 324/e Ans. three figures> 181, for the first par- 

tial dividend. 56 is contained in 
134 181 3 times, and a remainder; we 

112 write the 3 as the first figure in 

229 the quotient, and then multiply 

224 the divisor by this quotient figure ; 

^ 3 times 56 is 168, which subtracted 

from 181, leaves 13 ; to this re- 
mainder we annex or bring down 4, the next figure of the dividend, 
and thus form 134, the next partial dividend; 56 is contained in 
134 2 times, and a remainder; 2 times 56 is 112, which subtracted 
from 134, leaves 22; to this remainder we bring down 9, the last 
figure of the dividend, and we have 229, the last partial dividend ; 56 
is contained in 229 4 times, and a remainder ; 4 times 56' is 224, 
which subtracted from 229, gives 5, the final remainder, which we 
write in the quotient with the divisor below it, thus completing the 
division, (35). 

Note 2. — When the multiplication and subtraction are performed mentally, 
as in the first example, the operation is called Short Division; but when the 
work is written out in full, as in the second example, the operation is called 
Long Division. The principles governing the two methods are the same. 

109. From these principles and illustrations we derive the 
following general 

Rule. I. Beginning at the left hand, take for the first partial 
dividend the fewest figures of the given dividend that will contain 
the divisor one or more times ; find how many times the divisor is 
contained in this partial dividend, and write the result in the 
quotient; multiply the divisor by this quotient figure, and subtract 
the product from the partial dividend used. 



division. 49 

II. To the remainder bring down the next figure of the dividend, 
with which proceed as be/ore; and thus continue till all the figures 
of the dividend have been divided. 

III. If the division is not exact, place the final remainder in 
the quotient, and write the divisor underneath. 

110. Proof. There are two principal methods of proving 
division. 

1st. By multiplication. 

Multiply the divisor and quotient together, and to the product 

add the remainder, if any; if the result be equal to the dividend, 

the work is correct. (108, I.) 

Note. — la multiplication, the two factors are given to find the product ; in 
division, the product and one of the factors are given to find the other factor. 

2d. By excess of 9's. 

111. Subtract the remainder, if any, from the dividend, and 
find the excess of 9's in the result. Multiply the excess of 9's in 
the divisor by the excess of 9's in the quotient, and find the excess 
of 9's in the product; if the latter excess is the same as the 
former, the work is supposed to be correct. (85») 

EXAMPLES FOR PRACTICE. 

(1.) (2.) (3.) (4.) 
6)473832 8)972496 9)1370961 12)73042164 

Quotients. 

5. Divide 170352 by 36. 4732. 

6. Divide 409887 by 47. 8721. 

7. Divide 443520 by 84. 5280. 

8. Divide 36380250 by 125. 291042. 

9. Divide 1554768 by 216. 

10. Divide 3931476 by 556. 

11. Divide 48288058 by 3094. Bem. 

12. Divide 11214887 by 232. ... 7. 

13. Divide 27085946 by 216. 194. 

14. Divide 29137062 by 5317. 5219. 

15. Divide 4917968967 by 2359. 1255. 

5 d 



50 SIMPLE NUMBERS. 

16. What is the value of 721198 -*- 291 ? Kem. 100. 

17. What is the value of 3844449 -*- 657 ? 342. 

18. What is the value of 536819237 -r- 907 ? 403. 

19. What is the value of 571943007145 ~ 37149 ? 12214. 

20. What is the value of 48659910 -5- 54001 ? 5009. 

21. The annual receipts of a manufacturing company are 
$147675; how much is that per day, there being 365 days in the 
year? Ans. $404f if. 

22. The New York Central Railroad Company, in 1859, owned 
556 miles in length of railroad, which cost, for construction and 
equipment, $30732518 ; what was the average cost per mile ? 

^.$55,274111^ 

23. The Memphis and Charleston Railroad is 287 miles in 
length, and cost $5572470 ; what was the average cost per mile ? 

Ans. $19,416^. 

24. The whole number of Post offices in the United States, in 
1858, was 27977, and the revenue was $8186793 ; what was the 
average income to an office? 

ABBREVIATED LONG DIVISION. 

112. We may avoid writing the products in long division, 
and obtain the successive remainders by the method of subtraction 
employed in the case of several subtrahends. (76.) 

1. Divide 261249 by 487. 

operation. Analysis. Dividing the first partial 

487 ) 261249 (536 dividend, 2612, we obtain 5 for the first 

177 figure of the quotient. We now multi- 

313 ply 487 by 5 ; but instead of writing the 

217 Rem. product, and subtracting it from the 

partial dividend, we simply observe 

what figures must be added to the figures of the product, as we proceed, 

to give the figures of the partial dividend, and write them for the 

^mninder sought. Thus, 5 times 7 are 35, and 7 (written in the 

aaiiider,) are 42, a number whose unit figure is the same as the 

right hand figure of the partial dividend ; 5 times 8 are 40, and 4, the 

tens of the 42, are 44, and 7 (written in the remainder,) are 51 ; 5 






DIVISION. 51 

times 4 are 20, and 5, the tens of the 51, are 25, and 1 (written in the 
remainder,) are 26. We next consider the whole regnainder, 177, as 
joined with 4, the next figure of the dividend, making 1774 for the 
next partial dividend. Proceeding as before, we obtain 313 for the 
second remainder, 217 for the final remainder, and 536 for the entire 
quotient. Hence, the following 

Rule. L Obtain the first figure in the quotient in the usual 
manner. 

II. Multiply the first figure of the divisor by this quotient figure, 
and write such a figure in the remainder as, added to this partial 
product, will give an amount having for its unit figure the first or 
right hand figure of the partial dividend used. 

III. Carry the tens f figure of the amount to the product of the 
next figure of the divisor, and proceed as before, till the entire 
remainder is obtained. 

IV. Conceive this remainder to be joined to the next figure of 
the dividend, for a new partial dividend, and proceed as with the 
former, till the work is finished, 

EXAMPLES FOR PRACTICE. 



1. 


Divide 77112 by 204. 


Ans. 


378. 


2. 


Divide 65664 by 72. 


Ans. 


912. 


3. 


Divide 791B576 by 209. 


Ans. 


37864. 


4. 


Divide 6636584 by 698. 






5. 


Divide 4024156 by 8903. 


Arts. 


452. 


6. 


Divide 760592 by 6791. 






7. 


Divide 101443929 by 25203. 


Ans. 


4025 2 V 8 3 Vs 


8. 


Divide 1246038849 by 269181. 


Ans. 


4629. 


9. 


Divide 2318922 by 56240. 






10. 


Divide 1454900 by 17300. 


Ans. 


84 1700 



GENERAL PRINCIPLES OF DIVISION. 
1151. The general principles of division most important in their 
application, relate; 1st, to changing the terms of division by 
addition or subtraction; 2d, to changing the terms of division 
by multiplication or division ; 3d ; to successive division. 



g2 SIMPLE NUMBERS. 

114. The quotient in division depends upon the relative values 
of the dividend and divisor. Hence, any change in the value of 
either dividend or divisor must produce a change in the value of 
the quotient; though certain changes may be made in both divi- 
dend and divisor, at the same time, that will not affect the quotient. 

CHANGING THE TERMS BY ADDITION OR SUBTRACTION. 

U5. Since the dividend corresponds to a product, of which 
the divisor and quotient are factors, we observe, 

1st. If the divisor be increased by 1, the dividend must be 
increased by as many units as there are in the quotient, in order 
that the quotient may remain the same, (93, I) ; and if the divi- 
dend be not thus increased, the quotient will be diminished by as 
many units as the number of times the new divisor is contained 
in the quotient. Thus, 

84 -^ 6 = 14 

84 -v- 7 = 14 — V — 12 

2d. If the divisor be diminished by 1, the dividend must be 
diminished by as many units as there are in the quotient, in order 
that the quotient may remain the same, (03, II) ; and if the 
dividend be not thus diminished, the quotient will be increased 
by as many units as the number of times the new divisor is con- 
tained in the quotient. Thus, 

144 -- 9 = 16 

144 -- 8 = 16 + V 6 = 18 

These principles may be stated as follows : 

I. Adding 1 to the divisor takes as many units from the quotient 
as the new divisor is contained times in the quotient. 

II. Subtracting 1 from the divisor adds as many units to the 
quotient as the new divisor is contained times in the quotient. 
Hence, 

III. Adding any number to the divisor subtracts as many units 
from the quotient as the new divisor is contained times in the pro- 
duct of the quotient by the number added; and SUBTRACTING 



DIVISION. 53 

any number from the divisor adds as many units to the quotient 
as the new divisor is contained times in the product of the quo- 
tient by the number subtracted, 

CHANGING THE TERMS BY MULTIPLICATION OR DIVISION. 

116. There are six cases: 

1st. If any divisor is contained in a given dividend a certain 
number of times, the same divisor will be contained in twice the 
dividend twice as many times ; in three times the dividend, three 
times as many times ; and so on. Hence, 

Multiplying the dividend by any number, multiplies the quotient 
by the same number. 

2d. If any divisor is contained in a given dividend a certain 
number of times, the same divisor will be contained in one half 
the dividend one half as many times ; in one third the dividend, 
one third as many times ; and so on. Hence, 

Dividing the dividend by any number, divides the quotient by the 
same number. 

3d. If a given divisor is contained in any dividend a certain 
number of times, twice the divisor will be contained in the same 
dividend one half as many times ; three times the divisor, one 
third as many times ; and so on. Hence, 

Multiplying the divisor by any number, divides the quotient by 
the same number. 

4th. If a given divisor is contained in any dividend a certain 
number of times, one half the divisor will be contained in the 
same dividend twice as many times ; one third of the divisor, three 
times as many times ; and so on. Hence, 

Dividing the divisor by any number, multiplies the quotient by 
the same number. 

5th. If a given divisor is contained in a given dividend a cer- 
tain number of times, twice the divisor will be contained the same 
number of times in twice the dividend ; three times the divisor 
will be contained the same number of times in three times the 
dividend ; and so on. Hence, 
5* 



54 SIMPLE NUMBERS. 

Multiplying both dividend and divisor by the same number does 
not alter the quotient. 

6 th. If a given divisor is contained in a given dividend a cer- 
tain number of times, one half the divisor will be contained the 
same number of times in one half the dividend; one third of the 
divisor will be contained the same number of times in one third 
of the dividend ; and so on. Hence, 

Dividing both dividend and divisor by the same number does not 

alter the quotient 

Note. — If a number be multiplied and the product divided by the same num- 
ber, the quotient will be equal to the number multiplied; hence the 5th case 
may be regarded as a direct consequence of the 1st and 3d; and the 6th, as the 
direct consequence of the 2d and 4th. 

To illustrate these cases, take 24 for a dividend and 6 for a 
divisor ; then the quotient will be 4, and the several changes may- 
be represented in theii order as follows : 

Dividend. Divisor. Quotient. 

24 -+- 6 = 4 



1 48 _•_ fi R J Multiplying the dividend by 2 multi- 
1 plies the quotient by 2. 

2 12-i-fi 9 I Dividing the dividend by 2 divides 
( the quotient by 2. 

o 9/1 19 n (Multiplying the divisor by 2 divides 

6. ^ -f-lZ = ^j the quotient by 2. 

a 9 a o of Dividing the divisor by 2 multiplies 

4. Z<4 -fr- 6 — 8 J the (luotient by 2 . 

a ( Multiplying both dividend and divisor 
I by 2 does not alter the quotient. 

a ( Dividing both dividend and divisor by 
\ 2 does not alter the quotient. 

117. These six cases constitute three general principles, 
which may now be stated as follows : 

Prtn. I. Multiplying the dividend multiplies the quotient; and 
dividing the dividend divides the quotient. 

Prin. II. Multiplying the divisor divides the quotient; and 
dividing the divisor multiplies the quotient. 



5. 48 -*- 12 = 



6. 12 ~ 3 = 



division. 55 

Prin. III. Multiplying or dividing both dividend and divisor 
by the same number, does not alter the quotient. 

118* These three principles may be embraced in one 

GENERAL LAW. 

A change in the dividend produces a LIKE change in the quo- 
tient; but a change in the divisor produces an OPPOSITE change 
in the quotient. 

SUCCESSIVE DIVISION. 

119. Successive Division is the process of dividing one 
number by another, and the resulting quotient by a second divisor, 
and so on. 

Successive division is the reverse of continued multiplication. 
Hence, 

I. If a given number be divided by several numbers in succes- 
sive division, the result will be the same as if the given number 
were divided by the product of the several divisors, (95, I). 

II. The result of successive division is the same, in whatever 
order the divisors are taken, (95, II). 

CONTRACTIONS IN DIVISION, 
CASE I. 

ISO. When the divisor is a composite number. 
1. Divide 1242 by 54. 

Analysis. The component factors of 54 are 

OPERATION" 

6 and 9. We divide 1242 by 6, and the re- 

* suiting quotient by 9, and obtain for the final 

9)207 result, 23, which must be the same as the 

23 Ans. quotient of 1242 divided by 6 times 9, or 54, 

(119, I). We might have obtained the same 

result by dividing first by 9, and then by 6, (119, II). Hence the 

following 

Rule. Divide the dividend by one of the factors, and the quo* 



56 SIMPLE NUMBERS. 

tient thus obtained by another, and so on if there be more than two 
factors, until every factor has been made a divisor. The last quo- 
tient will be the quotient required. 



TO FIND THE TRUE REMAINDER. 

131. If remainders occur in successive division, it is evident 
that the true remainder must be the least number, which, sub- 
tracted from the given dividend, will render all the divisions 
exact 

I. Divide 5855 by 168, using the factors 3, 7, and 8, and find 
the true remainder, 

operation. Analysis. Dividing the 

8) 5855 given dividend by 3, we have 

7^1951 2 *^* ^ or a q u °ti ent > an d a 

remainder of 2. Hence, 2 

8)278 5x3= 15 subtracted from 5855 would 

34 ...6x 7x3=1 26 render the first division exact, 

True remainder 143 and we therefor e ^rite 2 for 

a part of the true remainder. 
Dividing 1951 by 7, we have 278 for a quotient, and a remainder of 5. 
Hence, 5 subtracted from 1951 would render the second division exact. 
But to diminish 1951 by 5 would require us to diminish 1951 X 3, the 
dividend of the first exact division, by 5 X 3 = 15, (93, III) ; and 
we therefore write 15 for the second part of the true remainder. 
Dividing 278 by 8, we have 34 for a quotient, and a remainder of 6. 
Hence, 6 subtracted from 278 would render the third division exact. 
But to diminish 278 by 6 would require us to diminish 278 X 7, the 
dividend of the second exact division, by 6 X 7 ; or 278 X 7 X 3, the 
dividend of the first exact division, by 6 X 7 X 3 = 126 ; and we 
therefore write 126 for the third part of the true remainder. Adding 
the three parts, we have 143 for the entire remainder. 

Hence the following 

Rule. I. Multiply each partial remainder by all the preceding 
divisors. 

II. Add the several products; the sum will be the true re- 
mainder. 



division. 57 

examples for practice. 

1. Divide 435 by 15 = 3 x 5. Ans. 29. 

2. Divide 4256 by 56 = 7 x 8. 

3. Divide 17856 by 72 = 9 x 8. 

4. Divide 15288 by 42 = 2 x 3 x 7. Ans. 364. 

5. Divide 972552 by 168 = 8 X 7 x 3. Ans. 5789. 

6. Divide 526050 by 126 = 9 X 7 X 2. 

7. Divide 612360 by 105 = 7 X 5 x 3. Ans. 5832. 

8. Divide 553 by 15 = 3 x 5. Rem. 13. 

9. Divide 10183 by 105 = 3 X 5 x 7. 103. 

10. Divide 10197 by 120 = 2 x 3 x 4 x 5. 117. 

11. Divide 29792 by 144 = 3 x 8 x 6. 128. 

12. Divide 73522 by 168 = 4 x 6 X 7. 106. 

13. Divide 63844 by 135 = 3 x 5 x 9. 124. 

14. Divide 386639 by 720 = 2 x 3 x 4 x 5 x 6. 719. 

15. Divide 734514 by 168 == 4 x 6 X 7. 18. 

16. Divide 636388 by 729 = 9 3 . 700. 

17. Divide 4619 by 125 = 5 s . 119. 

18. Divide 116423 by 10584 = 3 X 7 2 X 8 x 9. 10583. 

19. Divide 79500 by 6125 = 5 3 x 7 2 . 6000. 

CASE II. 

122. When the divisor is a unit of any order. 

If we cut off or remove the right hand figure of a number, each 
of the other figures is removed one place toward the right, and, 
consequently, the value of each is diminished tenfold, or divided 
by 10, (57 9 III). For a similar reason, by cutting off two figures 
.we divide by 100; by cutting off three, we divide by 1000, and 
so on ; and. the figures cut off will constitute the remainder. 
Hence the 

Rule. From the right hand of the dividend cut off as many 
figures as there are ciphers in the divisor. Under the figures so 
cut off j place the divisor , and the whole will form the Quotient. 



58 SIMPLE NUMBERS. 

EXAMPLES FOR PRACTICE. 

1. Divide 79 by 10. Ans. 7 T %. 

2. Divide 7982 by 100. 

3. Divide 4003 by 1000. Ans. 4-^. 

4. Divide 2301050 by 10000. 

5. Divide 3600036 by 1000. Ans. 8600 T j|fo. 

CASE III. 

123. "When there are ciphers on the right hand of 
the divisor. 

I. Divide 25548 by 700. 

Analysis. We resolve 700 

OPERATION. . , n _. , n 

into the factors 100 and 7. 

7|00) 255148 Dividing first by 100, the quo- 

36 Quotient. 3 2d rem. tient is 255, and the remainder 

3 X 100 + 48 = 348 true rem. 48. Dividing 255 by 7, the 

final quotient is 36, and the 
second remainder 3. Multiplying the last remainder, 3, by the 
preceding divisor, 100, and adding the preceding remainder, we have 
300+48 = 348, the true remainder, (121). In practice, the true 
remainder may be obtained by prefixing the second remainder to 
the first. Hence the 

Rule. I. Cut off the ciphers from the right of the divisor } and 
as many figures from the right of the dividend. 

II. Divide the remaining figures of the dividend by the remain- 
ing figures of the divisor y for the final quotient. 

III. Prefix the remainder to the figures cut off, and the result 
will be the true remainder. 

EXAMPLES FOR PRACTICE. 

1. Divide 7856 by 900. Ans. 8«* «. 

2. Divide 13872 by 500. 

3. Divide 83248 by 2600. Ans. 32^*^. 

4. Divide 1548036 by 4300. Ans. 360 ? |^. 

5. Divide 436000 by 300. Ans. 1453i{$. 

6. Divide 66472000 by 8100. 

7. Divide 10818000 by 3600. 



division. 59 



EXAMPLES COMBINING THE PRECEDING RULES. 

1. How many barrels of flour at $8 a barrel, will pay for 25 tons 
of coal at $4 a ton, and 36 cords of wood at $3 a cord ? 

Ans. 26. 

2. A grocer bought 12 barrels of sugar at $16 per barrel, and 
17 barrels at $13 per barrel ; how much would he gain by selling 
the whole at $18 per barrel ? 

3. A farmer sold 300 bushels of wheat at $2 a bushel, corn and 
oats to the amount of $750 ; with the proceeds he bought 120 
head of sheep at $3 a head, one pair of oxen for $90, and 25 acres 
of land for th^ remainder How much did the land cost him per 
acre ? Ans. 



4. Divide 450+ (24 — 12) x 5 by (90 -s- 6) + (3 x 11) — 18. 

Ans. 17. 

5. Divide 648 x (3 2 x 2 3 ) -s- 9 — (2910 ~- 15) by 2863 ~ 
(4375 -f- 175) X 4 2 + 3 2 ). Ans, 712f 

6. The product of three numbers is 107100; one of the 
numbers is 42, and another 34. What is the third number ? 

Ans. 75. 

7. What number is that which being divided by 45, the quo- 
tient increased by V + 1, the sum diminished by the difference 
between 28 and 16, the remainder multiplied by 6, and the pro- 
duct divided by 24, the quotient will be 12 ? Ans. 450. 

8. A mechanic earns $60 a month, but his necessary expenses 
are $42 a month. How long will it take him to pay for a farm 
of 50 acres worth $36 an acre ? 

9. What number besides 472 will divide 251104 without a re- 
mainder? Ans. 532. 

10. Of what number is 3042 both divisor and quotient ? 

Ans. 9253764. 

11. What must the number be which, divided by 453, will give 
the quotient 307, and the remainder 109 ? Ans. 139180. 

12. A farmer bought a lot of sheep and hogs, of each an equal 
number, for $1276. He gave $4 a head for the sheep, and $7 a 



gO SIMPLE NUMBERS. 

head for the hogs ; what was the whole number purchased, and 
how much was the difference in the total cost of each ? 

Arts. 116 purchased; $348 difference in cost. 

13. According to the census of 1850 the total value of the 
tobacco raised in the United States was $13,982,686. How many 
school-houses at a cost of $950, and churches at a cost of $7500, 
of each an equal number, could be built with the proceeds of the 
tobacco crop of 1850 ? Arts. 1654, and a remainder of $6386. 

14. The entire cotton crop in the United States in 1859 was 
4,300,000 bales, valued at $54 per bale. If the entire proceeds 
were exchanged for English iron, at $60 per ton, how many tons 
would be received ? 

15. The population of the United States in 1850 was 23,191,876. 
It was estimated that 1 person in every 400 died of intemperance. 
How many deaths may be attributed to this cause in the United 
States, during that year ? 

16. In 1850, there were in the State of New York, 10,593 
public schools, which were attended during the winter by 508464 
pupils ; what was the average number to each school ? 

Arts, 48. 

17. A drover bought a certain number of cattle for $9800, and 
sold a certain number of them for $7680, at $64 a head, and 
gained on those he sold $960. How much did he gain a head, 
and how many did he buy at first ? 

Ans. Gained $8 per head; bought 175. 

18. A house and lot valued at $1200, and 6 horses at $95 each, 
were exchanged for 30 acres of land. At how much was the land 
valued per acre ? 

19. If 16 men can perform a job of work in 36 days, in how 
many days can they perform the same job with the assistance of 
8 more men ? Ans. 24. 

20. Bought 275 barrels of flour for $1650, and sold 186 bar- 
rels of it at $9 a barrel, and the remainder for what it cost. How 
much was gained by the bargain ? Ans. $558. 

21. A grocer wishes to put 840 pounds of tea into three kinds 
of boxes, containing respectively 5, 10, and 15 pounds, putting the 



PROBLEMS. 61 

same number of pounds into each box. How many boxes can lie 
fill ? Ans. 28 of each kind. 

22. A coal dealer paid $965 for some coal. He sold 160 tons 
for $5 a ton, when the remainder stood him in but $3 a ton. How 
many tons did he buy ? Ans. 215. 

23. A dealer in horses gave $7560 for a certain number, and 
sold a part of them for $3825, at $85 each, and by so doing, lost 
$5 a head ; for how much a head must he sell the remainder, to 
gain $945 on the whole ? Ans. $120. 

24. Bought a Western farm for $22,360, and after expending 
$1742 in improvements upon it, I sold one half of it for $15480, 
at $18 per acre. How many acres did I purchase, and what did 
it cost me per acre ? 

PROBLEMS IN SIMPLE INTEGRAL NUMBERS. 

134:* The four operations that have now been considered, viz., 
Addition, Subtraction, Multiplication, and Division, are all the 
operations that can be performed upon numbers, and hence they 
are called the Fundamental Rules. 

125. In all cases, the numbers operated upon and the results 
obtained, sustain to each other the relation of a whole to its parts. 
Thus, 

I. In Addition, the numbers added are the parts, and the sum 
or amount is the whole. 
II. In Subtraction, the subtrahend and remainder are the 
parts, and the minuend is the whole. 

III. In Multiplication, the multiplicand denotes the value of one 

part, the multiplier the number of parts, and the pro- 
duct the total value of the whole number of parts. 

IV. In Division, the dividend denotes the total value of the 

whole number of parts, the divisor the value of one 
part, and the quotient the number of parts ; or the 
divisor the number of parts, and the quotient the 
value of one part. 

126. Every example that can possibly occur in Arithmetic, 
and every business computation requiring an arithmetical opera- 

6 



the greater number or the 
minuend. 



(52 SIMPLE NUMBERS. 

tion, can be classed under one or more of the four Fundamental 
Bules, as follows : 

L Cases requiring Addition. 
There may be given To find 

1. The parts, the whole, or the sum total. 

2 The less of two numbers and 
their difference, or the sub- 
trahend and remainder, 

II. Cases requiring Subtraction. 
There may be given To find 

1. The sum of two numbers and ") 

one of them, } the other - 

2. The greater and the less of ^ 

two numbers, or the minuend > the difference or remainder 
and subtrahend, J 

III. Cases requiring Multiplication. 
There may be given To find 

1. Two numbers, their product. 

2. Any number of factors, their continued product. 

3. The divisor and quotient, the dividend. 

IV. Cases requiring Division. 
There may be given To find 

1. The dividend and divisor, the quotient. 

2. The dividend and quotient, the divisor. 

3. The product and one of two ) . „ J 

n . y the other factor, 

factors, j 

4. The continued product of v 

several factors, and the pro- > that one factor, 
duct of all but one factor, J 

127. Let the pupil be required to illustrate the following pro- 
blems by original examples. 

Problem 1. Given, several numbers, to find their sum. 

Prob. 2^ Given, the sum of several numbers and all of them 
but one, to find that one. 



PROBLEMS. g3 

Prob. 3. Given, the parts, to find the whole. 

Prob. 4. Given, the whole and all the parts but one, to find 
that one. 

Prob. 5. Given, two numbers, to find their difference. 

Prob. 6. Given, the greater of two numbers and their difference, 
to find the less number. 

Prob. 7. Given, tfce less of two numbers and their difference, to 
find the greater number. 

Prob. 8. Given, the minuend and subtrahend, to find the 
remainder. 

Prob. 9. Given, the minuend and remainder, to find the sub- 
trahend. 

Prob. 10. Given, the subtrahend and remainder, to find the 
minuend. 

Prob. 11. Given, two or more numbers, to find their product. 

Prob. 12. Given, the product and one of two factors, to find the 
other factor. 

Prob. 13. Given, the continued product of several factors and 
all the factors but one, to find that factor. 

Prob. 14. Given, the factors, to find their product. 

Prob. 15. Given, the multiplicand and multiplier, to find the 
product. 

Prob. 16. Given, the product and multiplicand, to find the 
multiplier. 

Prob. 17. Given, the product and multiplier, to find the mul- 
tiplicand. 

Prob. 18. Given, two numbers, to find their quotients. 

Prob. 19. Given, the divisor and dividend, to find the quotient. 

Prob. 20. Given, the divisor and quotient, to find the dividend. 

Prob. 21. Given, the dividend and quotient, to find the divisor. 

Prob. 22. Given, the divisor, quotient, and remainder, to find 
the dividend. 

Prob. 23. Given, the dividend, quotient, and remainder, to find 
the divisor. 

Prob. 24. Given, the final quotient of a continued division and 
the several divisors, to find the dividend. 



64 SIMPLE NUMBERS. 

Prob* 25. Given, the final quotient of a continued division, the 
first dividend, and all the divisors but one, to find that divisor. 

Prob. 26. Given, the dividend and several divisors of a con- 
tinued division, to find the quotient. 

Prob. 27. Given, two or more sets of numbers, to find the 
difference of their sums. 

Prob. 28. Given, two or more sets of factors, to find the sum of 
their products. 

Prob. 29. Given, one or more sets of factors and one or more 
numbers, to find the sum of the products and the given numbers. 

Prob. 30. Given, two or more sets of factors, to find the differ- 
ence of their products. 

Prob. 3.1. Given / one or more sets of factors and one or more 
numbers, to find the sum of the products and the given number 
or numbers* 

Prob. 32. Given, two or more sets of factors and two or more 
other sets of factors, to find the difference of the sums of the 
former and latter. 

Prob. 33. Given, the sum and the difference of two numbers, to 
find the numbers. 

Analysis. If the difference of two unequal numbers be added to 
the less number, the sum will be equal to the greater ; and if this 
sum be added to the greater number, the result will be twice the 
greater number. But this result is the sum of the two numbers plus 
their difference. 

Again, if the difference of two numbers be subtracted from the 
greater number, the remainder will be equal to the less number ; and 
if this remainder be added to the less number, the result will be twice 
the less number. But this result is the sum of the two numbers 
minus their difference. Hence, 

I. The sum of two numbers plus their difference is equal to 
twice the greater number. 

II. The sum of two numbers minus their difference is equal to 
twice the less number. 



EXACT DIVISORS. 65 



PKOPEKTIES OP NUMBEKS. 

EXACT DIVISORS. 

128. An Exact Divisor of a number is one that gives an 
integral number for a quotient. And since division is the reverse 
of multiplication, it follows that all the exact divisors of a number 
are factors of that number, and that all its factors are exact 
divisors. 

Notes. — 1. Every number is divisible by itself and unity ; but the number 
itself and unity are not generally considered as factors, or exact divisors of the 
number. 

2. An exact divisor of a number is sometimes called the measure of the 
number. 

129 An Even Number is a number of which 2 is an exact 
divisor; as 2, 4, 6, or 8. 

130. An Odd Number is a number of which 2 is not an exact 
divisor; as 1, 3, 5, 7, or 9. 

131. A Perfect Number is one that is equal to the sum of 
all its factors plus 1; as 6 = 3 + 2 + 1, or 28 = 14 + 7 + 4 + 

2 + 1. 

Note.— The only perfect numbers known are 6, 28, 496,8128, 33550336, 
8589869056, 137438691328, 2305843008139952128, 2417851639228158837*84576, 
990352031428297183044881 6128 

133. An Imperfect Number is one that is not equal to the 
sum of all its factors plus 1 , as 12, which is not equal to 6 + 4 
+ 3 + 2 + 1. 

133. An Abundant Number is one which is less than the 
sum of all its factors plus 1 ; as 18, which is less than 9 + 6 + 

3 + 2 + 1. 

134. A Defective Number is one which is greater than the 
sum of all its factors plus 1 ; as 27, which is greater than 9+3 + 1. 

13o. To show the nature of exact division, and furnish tests 

of divisibility, observe that if we begin with any number, as 4, 

and take once 4, two times 4, three times 4, four times 4, and so 

on indefinitely, forming the series 4, 8, 12, 16, etc., we shall have 

6* E 



66 PROPERTIES OF NUMBERS. 

all the numbers that are divisible by 4 ; and from the manner of 
forming this series, it is evident, 

1st. That the product of any one number of the series by any 
integral number whatever, will contain 4 an exact number of 
times ; 

2d. The sum of any two numbers of the series will contain 4 
an exact number of times ; and 

3d. The difference of any two will contain 4 an exact number 
of times. Hence, 

I Any number which will exactly divide one of two numbers 
will divide their product. 

II. Any number which will exactly divide each of two numbers 
will divide their sum. 

III. Any number which will exactly divide each of two num- 
bers will divide their difference. 

130. From these principles we derive the following properties : 

I. Any number terminating with 0, 00, 000, etc., is divisible 
by 10, 100, 1000, etc., or by any factor of 10, 100, or 1000. 

For. by cutting off the cipher or ciphers, the number will be divided 
by 10, 100, or 1000, etc., without a remainder, (122) ; and a number 
of which 10, 100, or 1000, etc., is a factor, will contain any factor of 
10, 100, or 1000, etc., (I). 

II. A number is divisible by 2 if its right hand figure is even 
or divisible by 2. 

For, the part at the left of the units' place, taken alone, with its 
local value, is a number which terminates with a cipher, and is divi- 
sible by 2, because 2 is a factor of 10, (I) ; and if both parts, taken 
separately, with their local values, are divisible by 2, their sum, which 
is the entire number, is divisible by 2, (135, II). 

Note. — Hence, all numbers terminating with 0, 2, 4, 6, or 8, are even, and all 
numbers terminating with 1, 3, 5, 7, or 9, are odd, 

III. A number is divisible by 4 if the number expressed by its 
two right hand figures is divisible by 4. 

For, the part at the left of the tens' place, taken alone, with its 
local value, is a number which terminates with two ciphers, and is 
divisible by 4, because 4 is a factor of 100, (I) ; and if both parts, 



EXACT DIVISORS. g7 

taken separately, with their local values, are divisible by 4, their sum, 
which is the entire number, is divisible by 4, (135, II) 

IV. A number is divisible by 8 if the number expressed by is 
three right hand figures is divisible by 8. 

For, the part at the left of the hundreds' place, taken alone, with 
its local value, is a number which terminates with three ciphers, and 
is divisible by 8, because 8 is a factor of 1000, (I); and if both 
parts, taken separately, with their local values, are divisible by 8, 
their sum, or the entire number, *s divisible by 8, (135, II). 

V. A number is divisible by any power of 2, if as many right 
hand figures of the number as are equal to the index of the given 
power, are divisible by the given power. 

For, as 2 is a factor of 10, any power of 2 is a factor of the corres- 
ponding power of 10, or of a unit of an order one higher than is 
indicated by the index of the given power of 2 ; and if both parts 
of a number, taken separately, with their local values, are divisible 
by a power of 2, their sum, or the entire number, is divisible by the 
same power of 2, (135, II). 

VI. A number is divisible by 5 if its right hand figure is 0, 
or 5. 

For, if a number terminates with a cipher, it is divisible by 5, 
because 5 is a factor of 10, (I) ; and if it terminates with 5, both 
parts, the units and the figures at the left of units, taken separately, 
with their local values, are divisible by 5, and consequently their 
sum, or the entire number, is divisible by 5, (135, II). 

VII. A number is divisible by 25 if the number expressed by 
its two right hand figures is divisible by 25. 

For, the part at the left of the tens' figure, taken with its local 
value, is a number terminating with two ciphers, and is divisible by 
25, because 25 is a factor of 100, (I) ; and if both parts, taken 
separately, with their local values, are divisible by 25, their sum, or 
the entire number, is divisible by 25, (135, II). 

VIII. A number is divisible by any power of 5, if as many 
right hand figures of the number as are equal to the index of the 
given power are divisible by the given power. 

For, as 5 is a factor of 10, any power of 5 is a factor of the corres- 
ponding power of 10, or of a unit of an order one higher than is indi- 



gg PROPERTIES OF NUMBERS. 

cated by the index of the given power of 5 ; and if both parts of a 
number, taken separately, with their local values, are divisible by a 
power of 5, their sum, or the entire number, is divisible^by the same 
power of 5, (135, II). 

IX. A number is divisible by 9 if the sum of its digits is divis- 
ible by 9. 

For, if any number, as 7245, be separated into its parts, 7000 + 
200 + 40+5, and each part be divided by 9, the several remainders 
will be the digits 7, 2, 4, and 5, respectively ; hence, if the sum of 
these digits, or remainders, be 9 or an exact number of 9's, the entire 
number must contain an exact number of 9's, and will therefore be 
divisible by 9. 

Note. — "Whence it follows that if a number be divided by 9, the remainder 
•will be the same as the excess of 9's in the sum of the digits of the number. 
Upon this property depends one of the methods of proving the operations in 
the four Fundamental Rules. 

X. A number is divisible by a composite number, when it is 
divisible, successively, by all the component factors of the com- 
posite number. 

For, dividing any number successively by several factors, is the 
same as dividing by the product of these factors, (119, I). 

XI. An odd number is not divisible by an even number. 

For, the product of any even number by any odd number is even ; 
and, consequently, any composite odd number can contain only odd 
factors. 

XII. An even number that is divisible by an odd number, is 
also divisible by twice that odd number. 

For, if any even number be divided by an odd number, the quo- 
tient must be even, and divisible by 2 ; hence, the given even num- 
ber, being divisible successively by the odd number and 2, will be 
divisible by their product, or twice the odd number, (119, I). 

PRIME NUMBERS. 

137. A Prime Number is one that can not be resolved or 
separated into two or more integral factors. 

Note. — Every number must be either prime or composite. 






PRIME NUMBERS. £9 

138* To find all the prime numbers within any given limit, 
we observe that all even numbers except 2 are composite ; hence, 
the prime numbers must be sought among the odd numbers. 

139, If the odd numbers be written in their order, thus; 1, 
3, 5, 7, 9, 11, 13, 15 17, etc., we observe, 

1st. Taking every third number after 3, we have 3 times 3, 5 
times 3, 7 times 3, and so on ; which are the only odd numbers 
divisible by 3. 

2d. Taking every fifth number after 5, we have 3 times 5, 5 
times 5, 7 times 5, and so on; which are the only odd numbers 
divisible by 5. And the same will be true of every other number 
in the series. Hence, 

3d. If we cancel every third number, counting from 3, no 
number divisible by 3 will be left; and since 3 times 5 will be 
canceled, 5 times 5. or 25, will be the least composite number left 
in the series. Hence, 

4th. If we cancel every fifth number, counting from 25, no 
number divisible by 5 will be left; and since 3 times 7, and 5 
times 7, will be canceled, 7 4,imes 7, or 49, will be the least com- 
posite number left in the series. And thus with all the prime 
numbers. Hence, 

140. To find all the prime numbers within any given limit, 
we have the following 

Rule. I. Write all the odd numbers in their natural order. 

II. Cancel, or cross out, 3 times 3, or 9, and every third number 
after it; 5 times 5, or 25, and every fifth number after it; 7 times 
7, or 49, and every seventh number after it; and so on, beginning 
tvith the second power of each prime number in succession, till the 
given limit is reached. The numbers remaining, together tvith the 
number 2, will be the prime members required. 

Notes. — 1. It is unnecessary to count for every ninth number after 9 times 9, 
for being divisible by 3, they will be found already canceled; the same may be 
said of any other canceled, or composite number. 

2. This method of obtaining a list of the prime numbers was employed by 
Eratosthenes (born b. c, 275), and is called Eratosthenes' Sieve. 



70 



PROPERTIES OP NUMBERS. 
TABLE OF PRIME NUMBERS LESS THAN 1000. 



1 


59 


139 


233 


337 


439 


557 


653 


769 


883 


2 


61 


149 


239 


347 


443 


563 


659 


773 


887 


3 


67 


151 


241 


349 


449 


569 


661 


787 


907 


5 


71 


157 


251 


353 


457 


571 


673 


797 


911 


7 


73 


163 


257 


359 


461 


577 


677 


809 


919 


11 


79 


167 


263 


367 


463 


587 


683 


811 


929 


13 


83 


173 


269 


37a 


467 


593 


691 


821 


937 


17 


89 


179 


271 


379 


479 


599 


701 


823 


941 


19 


97 


181 


277 


383 


487 


601 


709 


827 


947 


23 


101 


191 


281 


389 


491 


607 


719 


829 


953 


29 


103 


193 


283 


397 


499 


613 


727 


839 


967 


31 


107 


197 


293 


401 


503 


617 


733 


853 


971 


37 


109 


199 


307 


409 


509 


619 


739 


857 


977 


41 


113 


211 


311 


419 


521 


631 


743 


859 


983 


43 


127 


223 


313 


421 


523 


641 


751 


863 


991 


47 


131 


227 


317 


431 


541 


643 


757 


877 


997 


53 


137 


229 


331 


433 


547 


647 


761 


881 








FACTORING. 
CASE I. 

141. To resolve any composite number into its 
prime factors. 

The Prime Factors of a number are those prime numbers 
which multiplied together will produce the given number. 

143. The process of factoring numbers depends upon the fol- 
lowing principles : 

I. Every prime factor of a number is an exact divisor of that 
nuniber. 

II. The only exact divisors of a number are its prime factors, or 
some combination of its prime factors. 

1. What are the prime factors of 798 ? 

Analysis. Since the given number is even, we 
divide by 2, and obtain an odd number, 399, for a 
quotient We then divide by the prime numbers 
3, 7, and 19, successively, and the last quotient is 
1. The divisors, 2, 3, 7, and 19, are the prime 
factors required, (II). Hence, the 



2 


798 


8 


399 


7 


133 


19 


19 



FACTORING. 71 

Rule. Divide the given number by any prime factor ; divide 
the quotient in the same manner , and so continue the division until 
the quotient is a prime number. The several divisors and the last 
quotient will be the prime factors required. 

Proof. The product of all the prime factors wil 1 be the given 
number. 

EXAMPLES FOR PRACTICE. 

1. What are the prime factors of 2150 ? 

2. What are the prime factors of 2447 ? 

3. What are the prime factors of 6300 ? 

4. What are the prime factors of 21504? 

5. What are the prime factors of 2366 ? 

6. What are the prime factors of 1000 ? 

7. What are the prime factors of 390625 ? 

8. What are the prime factors of 999999 ? 

143. If the prime factors of a number are small, as 2, 3, 5? 
7, or 11, they may be easily found by the tests of divisibility, 
(136), or by trial. But numbers may be proposed requiring many 
trials to find their prime factors. This difficulty is obviated, 
within a certain limit, by the Factor Table given on pages 72, 73. 

By prefixing each number in bold-face type in the column 
marked No., to the several numbers following it in the same divi- 
sion of the column, we shall form all the composite numbers less 
than 10,000, and not divisible by 2, 3, 5, 7, or 11 ; the numbers 
in the columns marked Fac, are the least prime factors of the 
numbers thus formed respectively. Thus, in one of the columns 
marked No., we find 39 in bold-face type, and below 39, in the 
same column, is 77, which annexed to 39, forms 3977, a composite 
number. The least prime factor of this number is 41, which we 
find at the right of 77, in the column marked Fac. 

14:41. Hence, for the use of this table, we have the following 

Rule. I. Cancel from the given number all factors less than 
I3 ; and then find the remaining factors by the table. 

II. If any number less than 10,000 is not found in the table, 
and is not divisible by 2, 5, 7, or 11, it is prime. 



72 



PROPERTIES OF NUMBERS. 



FACTOR TABLE. 



« 2 


<u 2 


4> £ 

•2 


2 


g - 
1 


« 2 

•2 


« 2 

•S 


•§ © 


« 2 

•9 © 


•9 


S » 

•° 


1 « 


a ~ 


a -5 

fc fa 




a z 


i « 

fc fa 


1 « 


1 » 


| 1 

& fa 


£ fa 


§ « 


1 


99 29 


11 17 


43 29 


79 37 


41 17 


83 17 


41 23 


09 31 


17 53 


77 31 


69 13 


9 


17 13 


57 19 


83 13 


51 13 


97 43 


59 17 


13 19 


27 29 


83 71 


3 


01 17 


57 31 


61 37 


89 19 


77 13 


34 


69 53 


21 29 


47 47 


91 29 


21 13 


23 13 


69 13 


63 13 


91 47 


83 19 


01 19 


87 13 


31 61 


59 67 


52 


47 13 


43 23 


15 


20 


25 


87 29 


03 41 


93 17 


43 43 


69 19 


07 41 


89 17 


49 13 


01 19 


21 43 


01 41 


93 41 


19 13 


39 


51 19 


71 13 


13 13 


99 13 


61 31 


13 17 


33 19 


07 23 


30 


27 23 


01 47 


69 17 


77 17 


19 17 


3 


89 23 


17 37 


41 13 


09 13 


07 31 


31 47 


37 31 


79 29 


48 


21 23 


23 17 


10 


41 23 


47 23 


33 17 


13 23 


39 19 


53 59 


81 13 


11 17 


39 13 


61 19 


03 17 


37 29 


59 29 


37 43 


29 13 


73 23 


59 37 


87 41 


19 61 


49 29 


77 13 


07 19 


77 19 


71 19 


61 13 


43 17 


81 59 


61 17 


93 23 


41 47 


51 59 


91 17 


27 13 


91 37 


77 31 


67 17 


53 43 


97 13 


73 29 


99 53 


43 29 


63 19 


4 


37 17 


16 


21 


73 31 


71 37 


35 


77 41 


44 


47 37 


67 23 


03 13 


73 29 


33 23 


17 29 


81 29 


77 17 


03 31 


79 23 


27 19 


49 13 


87 17 


37 19 


79 13 


43 31 


19 13 


87 13 


97 19 


23 13 


91 13 


29 4 3 


53 23 


93 61 


81 13 


81 23 


49 17 


47 19 


99 23 


31 


51 53 


40 


39 23 


5943 


53 


93 17 


11 


51 13 


59 17 


26 


03 29 


69 43 


09 19 


53 61 


67 31 


11 47 


5 


21 19 


79 23 


71 13 


03 19 


07 13 


87 17 


31 29 


69 41 


83 19 


17 13 


27 17 


39 17 


81 41 


73 41 


23 4 3 


27 53 


89 37 


33 37 


71 17 


91 67 


21 17 


29 23 


47 3* 


91 19 


83 37 


27 37 


31 31 


99 59 


43 13 


89 67 


97 59 


29 7 3 


33 13 


57 13 


IT 


97 13 


41 19 


33 13 


36 


61 31 


45 


49 


39 19 


51 19 


59 19 


03 13 


22 


69 17 


39 43 


01 13 


63 17 


11 13 


01 13 


53 53 


59 13 


89 29 


11 29 


01 31 


27 


49 47 


11 23 


69 13 


31 23 


13 17 


59 23 


89 19 


13 


17 17 


09 47 


01 37 


51 23 


29 19 


87 61 


37 13 


27 13 


63 31 


6 


07 17 


39 37 


27 17 


43 13 


61 29 


49 41 


97 17 


41 19 


79 13 


71 41 


11 13 


19 23 


51 17 


31 23 


47 41 


73 19 


53 13 


41 


53 29 


81 17 


77 19 


29 17 


41 17 


63 41 


49 13 


59 31 


93 31 


67 19 


17 23 


59 47 


97 19 


89 17 


67 23 


47 29 


69 29 


57 37 


71 17 


97 23 


79 13 


21 13 


73 17 


50 


54 


89 13 


61 13 


81 13 


63 31 


73 47 


32 


83 29 


41 41 


77 23 


17 29 


29 61 


97 17 


71 31 


18 


79 43 


28 


U 13 


37 


63 23 


79 19 


29 47 


47 13 


7 


73 19 


07 13 


91 29 


09 53 


33 53 


13 47 


71 43 


89 13 


41 71 


5947 


03 19 


13 


17 23 


23 


13 29 


39 41 


21 61 


81 37 


46 


53 31 


61 43 


13 23 


13 13 


19 17 


23 23 


31 19 


47 17 


37 37 


83 4 7 


01 43 


57 13 


73 13 


31 17 


33 31 


29 31 


27 13 


39 17 


63 13 


43 19 


87 53 


07 17 


63 61 


91 17 


67 13 


39 13 


43 19 


29 17 


67 47 


77 29 


49 23 


89 59 


19 31 


69 37 


97 23 


79 19 


43 17 


49 43 


53 13 


69 19 


81 17 


57 13 


99 13 


33 41 


83 13 


55 


93 13 


49 19 


53 17 


63 17 


73 13 


87 19 


63 53 


42 


61 59 


51 


13 37 


99 17 


57 23 


91 31 


69 23 


81 43 


93 37 


81 19 


23 41 


67 13 


11 19 


39 29 


8 


63 29 


19 


24 


99 13 


33 


91 17 


37 19 


81 31 


23 4 7 


43 23 


17 19 


69 37 


09 23 


07 29 


29 


17 31 


99 29 


47 31 


87 43 


29 23 


49 31 


41 29 


87 19 


19 19 


13 19 


11 41 


37 47 


38 


67 17 93 13 


41 53 


61 67 


51 23 


91 13 


21 17 


19 41 


21 23 


41 13 


09 13 


43 


99 37 


43 37 


67 19 


71 13 


11 


27 41 


49 31 


23 37 


49 17 


11 37 


03 13 


47 


49 19 


87 37 


93 19 


03 23 


37 13 


61 23 


29 29 


79 31 


27 43 


07 59 


07 17 


61 13 


97 29 



FACTORING, 



73 



FACTOR TABLE — Continued. 



£ * 
_§ £ 

1 1 


2 £ 

s I 


1 1 


£ « 

1 s 


s £ 

i 8 


II 


« £ 

1 s 


£ -• 

<a £ 

1 1 

3 * 


50 
6 2 


1 1 


» £ 

a ° 
3 % 


S3 Ph 


£ fe 


S3 In 


» c« 


S3 In 


S3 In 


S3 Cm 


S3 In 


S3 Ix 


S3 £ 


S3 (x 


56 


60 


39 47 


51 13 


61 53 


57 13 


23 71 


13 47 


51 53 


59 43 


73 17 


03 13 


01 17 


43 17 


59 19 


67 13 


61 47 


27 23 


17 19 


57 17 


63 59 


83 23 


09 71 


19 13 


63 23 


77 13 


77 19 


63 79 


33 29 


41 23 


73 19 


69 13 


97 


11 31 


23 19 


67 29 


87 71 


79 29 


97 43 


47 13 


53 79 


79 13 


71 73 


01 89 


17 41 


31 37 


87 13 


89 83 


89 37 


77 


51 83 


71 43 


81 83 


87 37 


03 31 


27 17 


49 23 


93 43 


93 61 


91 23 


09 13 


77 41 


73 37 


91 17 


99 17 


07 17 


29 13 


59 73 


97 73 


69 


73 


29 59 


83 59 


79 61 


89 


93 


27 71 


33 43 


71 13 


99 67 


01 67 


03 67 


39 71 


81 


83 17 


03 29 


01 67 


31 37 


71 53 


77 53 


65 


13 31 


13 71 


47 61 


19 23 


89 13 


09 59 


07 41 


61 43 


81 13 


61 


09 23 


29 13 


19 13 


51 23 


31 47 


97 29 


17 37 


13 59 


63 13 


99 41 


03 17 


11 17 


31 29 


27 17 


69 17 


37 79 


85 


27 79 


29 19 


73 29 


57 


07 31 


27 61 


43 53 


39 41 


77 19 


43 17 


07 47 


47 23 


47 13 


97 97 


07 13 


09 41 


3347 


53 17 


61 17 


81 31 


49 29 


09 67 


57 13 


5343 


99 41 


13 29 


19 29 


39 13 


73 19 


63 37 


83 43 


53 31 


31 19 


59 17 


67 17 


98 


23 59 


37 17 


41 31 


89 29 


67 53 


87 13 


59 41 


49 83 


77 47 


79 83 


09 17 


29 17 


57 47 


57 79 


TO 


73 73 


78 


77 13 


51 17 


83 13 


89 41 


27 31 


59 13 


61 61 


83 29 


03 4 7 


7947 


01 29 


89 19 


57 43 


89 89 


94: 


41 13 


67 73 


69 31 


93 19 


09 43 


87 83 


07 37 


83 


67 13 


93 17 


07 23 


47 43 


71 29 
73 23 


79 37 


66 


31 79 


91 19 


11 73 


01 59 

03 13 


79 23 


90 


09 97 
51 13 


53 59 

69 71 


87 23 


13 17 


33 13 


97 13 


13 13 


87 31 


17 71 


77 53 


91 41 


17 13 


37 31 


74: 


31 41 


07 29 


93 13 


19 29 


69 17 


81 41 


58 


62 


23 37 


«1 23 


09 31 


37 17 


13 4 3 


86 


47 83 


81 19 


93 13 


09 37 


27 13 


31 19 


67 37 


21 41 


49 47 


27 19 


11 79 


71 47 


87 53 


99 19 


33 19 


33 23 


41 29 


81 73 


23 13 


59 29 


49 73 


21 37 


73 43 


95 


99 


37 13 


39 17 


47 17 


87 19 


29 17 


71 17 


51 37 


33 89 


77 29 


09 37 


13 23 


91 43 


41 79 


49 61 


93 41 


39 43 


91 13 


57 23 


39 53 


83 31 


17 31 


17 47 


93 71 


53 13 


67 59 


97 47 


53 29 


97 53 


79 17 


51 41 


89 61 


23 89 


37 19 


99 17 


83 61 


83 41 


99 31. 


63 17 


79 


99 43 


53 17 


91 


29 13 


43 61 


59 


89 19 


97 37 


71 


71 31 


13 41 


83 


71 13 


01 19 


53 41 


53 37 


09 19 


63 


67 


It 13 


93 59 


21 89 


03 19 


83 19 


13 13 


57 19 


59 23 


11 23 


13 59 


07 19 


23 17 


75 


39 17 


21 53 


87 


31 23 


63 71 


71 13 


17 61 


19 71 


31 53 


41 37 


01 13 


43 13 


33 13 


11 31 


39 13 


71 17 


79 17 


21 31 


31 13 


39 23 


53 23 


19 73 


57 73 


39 31 


17 2? 


43 37 


77 59 


91 97 


33 17 


41 17 


49 17 


57. 17 


31 17 


61 19 


41 19 


49 13 


67 89 


89 43 


97 19 


41 13 


71 23 


51 43 


63 13 


43 19 


67 31 


43 17 


59 19 


69 53 


93 53 




47 19 


83 13 


57 29 


69 67 


71 67 


69 13 


57 61 


73 31 


79 67 


99 29 




59 59 


64 


67 67 


71 71 


97 71 


79 79 


59 13 


77 67 


93 29 


96 




63 67 


01 37 


73 13 


81 43 


76 


81 23 


81 17 


91 59 


97 17 


07 13 




69 47 


03 19 


99 13 


97 23 


13 23 


91 61 


83 83 


97 19 


93 


17 59 




83 31 


07 4 3 


68 


73. 


19 19 


99 19 


99 37 


88 


11 61 


37 23 




89 53 


09 13 


17 17 


01 19 


27 29 


80 


84 


01 13 


17 13 


41 31 




93 13 


31 59 


21 19 


23 31 


31 13 


03 53 


01 31 


09 23 


23 23 


59 13 




97 43 


37 41 


47 41 


41 13 


33 17 


21 13 


11 13 


43 37 


53 19 


71 19 




















_ 







3 


188139 


7 


62713 


17 


8959 


17 


527 




31 



74 PROPERTIES OF NUMBERS. 

1. Resolve 1961 into its prime factors. 

operation. Analysis. Cutting off the two 

1961 -~ 37 = 53 right hand figures of the given 

1961 __ 37 x 53, Arts. number, and referring to the table, 

column No., we find the other part, 
19, in bold-face type ; and under it, in the same division of the column, 
we find 61, the figures cut off; at the right of 61, in column Fac, we 
find 37, the least prime factor of the given number. Dividing by 37, 
we obtain 53, the other factor. 

2. Resolve 188139 into its prime factors. 

operation. Analysis. We find by trial 

that the given number is divisible 
by 3 and 7 ; dividing by these fac- 
tors, we have for a quotient 8959. 
By referring to the factor table, 
we find the least prime factor of 
this number to be 17 ; dividing by 
17, we have 527 for a quotient, 
3 X 7 X 17 X 17 X 31, Ans. Keferring again to the table, we 

find 17 to be the least factor of 
527, and the other factor, 31, is prime. 

EXAMPLES FOR PRACTICE. 

1. Resolve 18902 into its prime factors. Ans. 2 ; 13, 727. 

2. Resolve 352002 into its prime factors. 

3. Resolve 6851 into its prime factors. 

4. Resolve 9367 into its prime factors. 

5. Resolve 203566 into its prime factors. 

6. Resolve 59843 into its prime factors. 

7. Resolve 9991 into its prime factors. 

8. Resolve 123015 into its prime factors. 

9. Resolve 893235 into its prime factors. 

10. Resolve 390976 into its prime factors* 

11. Resolve 225071 into its prime factors. 

12. Resolve 81770 into its prime factors. 

13. Resolve 6409 into its prime factors. 

14. Resolve 178296 into its prime factors. 

15. Resolve 714210 into its prime factors. 






FACTORING. 75 

CASE II. 

145. To find all the exact divisors of a number. 

It is evident that all the prime factors of a number, together 
with all the possible combinations of those prime factors, will con- 
stitute all the exact divisors of that number, (142, II). 

1. What are all the exact divisors of 860 ? 

OPERATION. 

60=lx2x2x 2x3x3x5. 

1,2, 4 , 8 Combinations of 1 and 2. 

9; 18; 36 ; n] " "Iand2and3. 

5 , 10 , 20 , 40 " "1 and 2 and 5. 

15 , 30 , 60 , 120 ) ,, „ , , „ , „ , , 

45 , 90 J 180 J 360 j X and 2 and 3 and 5 - 

Analysis. By Case I we find the prime factors of 360 to be 1, 2, 
2, 2, 3, 3, and 5. As 2 occurs three times as a factor, the different 
combinations of 1 and 2 by which 360 is divisible will be 1, 1x2 = 2, 
1 X 2 X 2 = 4, and 1x2x2 X 2 = 8; these we write in the first line. 
Multiplying the first line by 3 and writing the products in the second 
line, and the second line by 3, writing the products in the third line, 
we have in the first, second and third lines all the different combina- 
tions of 1, 2, and 3, by which 360 is divisible. Multiplying the first, 
second and third lines by 5, and writing the products in the fourth, 
fifth and sixth lines, respectively, we have in the six lines together, 
every combination of the prime factors by which the given number, 
360, is divisible. 

Hence the following 

Rule. I. Resolve the given number into its prime factors. 

II Form a series having 1 for the firstderm, that prime factor 
which occurs the greatest number of times in the given number for 
the second term, the square of this factor for the third term, and so 
on, till a term is reached containing this factor as many times as it 
occurs in the given number. 

III. Multiply the numbers in this line by another factor, and 
these results by the same factor, and so on, as many times as this 
factor occurs in the given number. 



76 PROPERTIES OF NUMBERS. 

IV. Multiply all the combinations now obtained by another 
factor in continued multiplication, and thus proceed till all the dif- 
ferent factors have been used. All the combinations obtained will 
be the exact divisors sought 



EXAMPLES FOR PRACTICE. 

1. What are all the exact divisors of 120 ? 

Ans. 1, 2, 3,4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. 

2. Find all the exact divisors of 84. 

Am. 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84. 

3. Find all the exact divisors of 100. 

Ans. 1, 2, 4, 5, 10, 20, 25, 50, 100. 

4. Find all the exact divisors of 420. 

Am f 1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 
' 1 30, 35, 42, 60, 70, 84, 105. 140, 210, 420. 

5. Find all the exact divisors of 1050. 
, ( 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 25, 30, 35, 42, 50, 70, 

' X 75, 105, 150, 17" 






175, 210, 350, 525, 1050. 

GREATEST COMMON DIVISOR. 

1 10. A Common Divisor of two or more numbers is a number 
that will exactly divide each of them. 

147. The Greatest Common Divisor of two or more numbers 
is the greatest number that will exactly divide each of them. 

148. Numbers Prime to each other are such as have no com- 
mon divisor. 

Note. — A common divisor is sometimes called a Common Measure ; and the 
greatest common divisor, the Greatest Common Measure. 

CASE I. 

149. When the numbers can be readily factored. 

It is evident that if several numbers have a common divisor, 
they may all be divided by any component factor of this divisor, 
and the resulting quotients by another component factor, and so 
on, till all the component factors have been used. 



GREATEST COMMON DIVISOR. 77 

1. What is the greatest common divisor of 28, 140, and 420 ? 
operation. Analysis. We readily see that 7 

28 . . 140 . . 420 will exactly divide each of the given 
"7 on p~t\ numbers; and then, 4 will exactly 

— — divide each of the resulting quotients. 

15 Hence, each of the given numbers 



4 X 7 = 28, Ans. can be exactly divided by 7 times 4 ; 
and these numbers must be compo- 
nent factors of the greatest common divisor. Now, if there were any 
other component factor of the greatest common divisor, the quotients, 
1, 5 and 15, would be divisible by it. But these quotients are prime 
to each other ; therefore, 7 and 4 are all the component factors of the 
greatest common divisor sought. 

From this analysis we derive the following 

Rule. I. Write the numbers in a line, with a vertical line at 
the lefty and divide by any factor common to all the numbers. 

II. Divide the quotients in like manner, and continue the divi- 
sion till a set of quotients is obtained that are prime to each other. 

II I. Multiply all the divisors together, and the product will be 
the greatest common divisor sought 

EXAMPLES FOR PRACTICE. 

1. What is the greatest common divisor of 40, 75, and 100? 

Ans. 5. 

2. What is the greatest common divisor of 18, 30, 36, 42, 
and 54 ? 

3. What is the .greatest common divisor of 42, 63, 126, and 
189? Ans. 21. 

4. What is the greatest common divisor of 135, 225, 270, and 
315? Ans. 45. 

5. What is the greatest common divisor of 84, 126, 210, 252, 
294, and 462 ? 

6. What is the greatest common divisor of 216, 360, 432, 648, 
and 936 ? Ans. 72. 

7. What is the greatest common divisor of 102, 153, and 255 ? 

Ans. 51. 
7* 



78 PROPERTIES OF NUMBERS. 

8. What is the greatest common divisor of 756, and 1575 ? 

9. What is the greatest common divisor of 182, 364, and 455? 

10. What is the greatest common divisor of 2520, and 3240 ? 

Ans. 360. 

11. What is the greatest common divisor of 1428, and 1092 ? 

12. What is the greatest common divisor of 1008, and 1036 ? 

Ans. 28. 

CASE II. 

ISO. When the numbers cannot be readily factored. 

The analysis of the method in this case depends upon the 
following properties of divisors. 

I. An exact divisor divides any number of times its dividend. 

II. A common divisor of two numbers is an exact divisor of 
their sum. 

III. A common divisor of two numbers is an exact divisor of 
their difference. 

Note. — The last two properties are essentially the same as 102, II, III. 

1. What is the greatest common divisor of 527, and 1207? 
operation. Analysis. We will first describe the pro- 

cess, and then examine the reasons for the 
several steps in the operation. Drawing two 
vertical lines, we place the greater number 
on the right, and the less number on the left, 
one line lower down. We then divide 1207, 
the greater number, by 527, the less, and 
write the quotient, 2, between the verticals, 
the product, 1054, opposite the less number and under the greater, 
and the remainder, 153, below. We next divide 527 by this re- 
mainder, writing the quotient, 3, between the verticals, the product, 
459, on the left, and the remainder, 68, below. We again divide the 
last divisor, 153, by 68, and obtain 2 for a quotient, 136 for a pro- 
duct, and 17 for a remainder, all of which we write in the same order 
as in the former steps. Finally, dividing the last divisor, 68, by the 
last remainder, 17, we have no remainder, and 17, the last divisor, is 
the greatest common divisor of the given numbers. 

Now, observing that the dividend is always the sum of the product 
and remainder, and that the remainder is always the difference of the 



527 
459 



68 
68 





1207 


2 


1054 


3 


153 


2 
4 


136 
17 



GREATEST COMMON DIVISOR. 



79 



dividend and product, we first trace the work in the reverse order, as 
indicated by the arrow line in the diagram below. 

17 divides 68, as proved by the 



OPERATION. 



527 



459 



68 



68 





i 

2 


3 


2 




* ^ 



1207 



1054 



153 



136 



17 



last division ; it will also divide 
2 times 68, or 136, (I). Now, as 
17 divides both itself and 136, it 
will divide 153, their sum, (II). 
It will also divide 3 times 153, or 
459, (I) ; and since it is a com- 
mon divisor of 459 and 68, it 
must divide their sum, 527, which 
is one of the given numbers. It 
will also divide 2 times 527, or 



1054, (I) ; and since it is a common divisor of 1054 and 153, it must 
divide their sum, 1207, the greater number, (II). Hence, 17 is a, com- 



mon divisor of the given numbers. 



Again, tracing the work in the direct order, as indicated in the fol- 
lowing diagram, we know that 



527 m^- 



459 



68 



1207 



1054 



153 



136 



17 



the greatest common divisor, what- 
ever it be, must divide 2 times 
527, or 1054, (I). And since it 
will divide both 1054 and 1207, 
it must divide their difference, 
153, (III). It will also divide 3 
times 153, or 459, (I) ; and as it 
will divide both 459 and 527, it 
must divide their difference, 68, 
(III). It will also divide 2 times 
68, or 136, (I) ; and as it will 

divide both 136 and 153, it must divide their difference, 17, (III) ; 

hence, it cannot be greater than 17. 

Thus we have shown, 

1st. That 17 is a common divisor of the given numbers. 
2d. That their greatest common divisor, whatever it be, cannot 
be greater than 17. Hence it must be 17. 

From this example and analysis, we derive the following 

Rule. I. Draw two verticals, and write the two numbers, one 
on each side, the greater number one line above the less. 



80 



PROPERTIES OF NUMBERS. 



II. Divide the greater number by the less, writing the quotient 
between the verticals, the product under the dividend, and the re- 
mainder below. 

III. Divide the less number by the remainder, the last divisor 
by the last remainder, and so on, till nothing remains. The last 
divisor will be the greatest common divisor sought. 

IV. If more than two numbers be given, first find the greatest 
common divisor of two of them, and then of this divisor and one 
of the remaining numbers, and so on to the last ; the last common 
divisor found will be the greatest common divisor required. 

Notes. — 1. When more than two numbers are given, it is better to begin with 
the least two. 

2. If at any point in the operation a prime number occur as a remainder, it 
must be a common divisor, or the given numbers have no common divisor. 



EXAMPLES FOR PRACTICE. 

1. What is the greatest common divisor of 18607 and 417979? 

OPERATION. 

417979 



18607 



, 17250 



1357 

966 



391 
368 



Ans. 



23 



2 


37214 




45839 


2 


37214 


2 


8625 


6 


8142 


2 


483 


1 


391 


4 


92 


4 


92 






2. What is the greatest common divisor of 10661 and 12303? 



OPERATION. 

12303 



10661 

9852 



Prime 809 



10661 



1642 



Ans. 1. 



GREATEST COMMON DIVISOR. gl 

3. What is the greatest common divisor of 336 and 812 ? 

Ans. 28. 

4. What is the greatest common divisor of 407 and 1067 ? 

5. What is the greatest common divisor of 825 and 1372 ? 

6. What is the greatest common di~ : sor of 2041 and 8476 ? 

Ans. 13. 

7. What is the greatest common divisor of 3281 and 10778 ? 

8. Find the greatest common divisor of 22579, and 116939. 

9. What is the greatest common divisor of 49373 and 147731 ? 

Ans. 97. 

10. What is the greatest common divisor of 1005973 and 
4616175 ? 

11. Find the greatest common divisor of 292, 1020 ; and 1095. 

Ans. 1. 

12. What is the greatest common divisor of 4718, 6951, and 
8876? Ans. 7. 

13. Find the greatest common divisor of 141, 799, and 940. 

14. What is the greatest common divisor of 484391 and 684877 ? 

Ans. 701. 

15. A farmer wishes to put 364 bushels of corn and 455 bushels 
of oats into the least number of bins possible, that shall contain 
the same number of bushels without mixing the two kinds of 
grain ; what number of bushels must each bin hold ? 

Ans. 91. 

16. A gentleman having a triangular piece of land, the sides of 
which are 165 feet, 231 feet, and 385 feet, wishes to inclose it 
with a fence having pannels of the greatest possible uniform 
length; what will be the length of each pannel? 

17. B has $620, C $1116, and D $1488, with which they agree 
to purchase horses, at the highest price per head that will allow 
each man to invest all his money ; how many horses can each man 
purchase ? Ans. B 5, C 9, and D 12. 

18. How many rails will inclose a field 14599 feet long by 
10361 feet wide, provided the fence is straight, and 7 rails high ; 
and the rails of equal length, and the longest that can be used ? 

Ans, 26880. 

F 



g2 PROPERTIES OF NUMBERS. 



, 



LEAST COMMON MULTIPLE. 

151. A Multiple is a number exactly divisible by a given 

number ; thus, 20 is a multiple of 4. 

Notes. — 1. A multiple is necessarily composite; a divisor may be either 
prime or composite. 

2. A number is a divisor of all its multiples and a multiple of all its divisors. 

153. A Common Multiple is a number exactly divisible by 
two or more grven numbers ; thus, 20 is a common multiple of 2, 
4, 5, and 10. 

153. The Least Common Multiple of two or more numbers 
is the least number exactly divisible by those numbers ; thus, 24 
is the least common multiple of 3, 4, 6, and 8. 

154:. From the definition it is evident that the product of two 
or more numbers, or any number of times their product, must be 
a common multiple of the numbers. Hence, A common multiple 
of two or more numbers may be found by multiplying the given 
numbers together. 

155. To find the least common multiple. 

FIRST METHOD. 

From the relations of multiple and divisor we have the following 
properties : 

I. A multiple of a number must contain all the prime factors 
of that number. 

II. A common multiple of two or more numbers must contain 
all the prime factors of each of those numbers. 

III. The least common multiple of two or more numbers must 
contain all the prime factors of each of those numbers, and no 
other factors. 

1. Find the least common multiple of 63, 66, and 78. 

operation. Analysis. The 

63 = 3 X 3 X 7 number cannot be less 

66 = 2 X 3 X 11 than 78, because it 

78 = 2 X 3 X 13 must contain 78 ; and 

2x3x 13 X 11x3x7 = 18018 Ans. if it contains 78, it 

must contain all its 
prime factors, viz. ; 2 X 3 X 13. 



LEAST COMMON MULTIPLE. g3 

We here have all the prime factors, and also all the factors of 66 
except 11. Annexing 11 to the series of factors, 

2 X 3 X 13 X 11, 
and we have all the prime factors of 78 and 66, and also all the fac- 
tors of 63 except one 3, and 7. Annexing 3 and 7 to the series of 

factors, 

2 X 3 X 13 X 11 X 3 X 7, 
and we have all the prime factors of each of the given numbers, and 
no others; hence the product of this series of factors is the least 
common multiple of the given numbers, (III). 

From this example and analysis we deduce the following 
Rule. I. Resolve the given numbers into their prime factors. 
II. Multiply together all the prime factors of the largest number } 
and such prime factors of the other numbers as are not found in 
the largest number, and their product will be the least common 
multiple. 

Note. — When a prime factor is repeated in any of the given numbers, it 
must be taken as many times in the multiple, as the greatest number of times it 
appears in any of the given numbers. 

EXAMPLES FOR PRACTICE. 

1. Find the least common multiple of 60, 84, and 132. 

Ans. 4620. 

2. Find the least common multiple of 21, 30, 44, and 126. 

Ans. 13,860. 

3. Find the least common multiple of 8, 12, 20, and 30. 

4. Find the least common multiple of 16, 60, 140, and 210. 

Am. 1,680. 

5. Find the least common multiple of 7, 15, 21, 25, and 35. 

6. Find the least common multiple of 14, 19, 38, 42, and 57. 

Ans. 798. 

7. Find the least common multiple of 144, 240, 480, 960. 

SECOND METHOD. 

150. 1* What is the least common multiple of 4, 9, 12, 18, 

and 36? 



84 



PROPERTIES OP NUMBERS. 



2 


4. 


UJTJ 

.9. 


JKAT1UI 

. 12. 


.18. 
. 9 . 


.36 


2 


2. 


.9. 


. 6. 


.18 


3 




9. 


. 3 . 


. 9. 


. 9 


3 




3 


3 




3 



2x2x3x3 = 36 Ans. 



Analysis. We first write 
the given numbers in a se- 
ries with a vertical line at 
the left. Since 2 is a fac- 
tor of some of the given 
numbers, it must be a factor 
of the least common mul- 
tiple sought, (155, II). Di- 



viding as many of the numbers as are divisible by 2, we write the 
quotients, and the undivided number, 9, in a line underneath. Now, 
since some of the numbers in the second line contain the factor 2, the 
least common multiple must contain another 2, and we again divide 
by 2, omitting to write any quotient when it is 1. "We next divide 
by 3 for a like reason, and still again by 3. By this process we have 
transferred all the factors of each of the numbers to the left of the 
vertical ; and their product, 36, must be the least common multiple 
sought, (155, III). 

2. What is the least common multiple of 20, 12, 15, and 75 ? 

Analysis. We readily 
see that 2 and 5 are among 
the factors of the given num- 
bers, and must be factors of 
the least common multiple ; 
hence, writing 2 and 5 at the 
left, we divide every number 
that is divisible by either of these factors or by their product ; thus, 
we divide 20 by both 2 and 5 ; 12 by 2 ; 15 by 5 ; and 75 by 5. We 
next divide the second li'ne in like manner by 2 and 3 ; and afterward 
the third line by 5. By this process we collect the factors of the 
given numbers into groups ; and the product of the factors at the 
left of the vertical is the least common multiple sought. 

3. What is the least common multiple of 7, 10, 15, 42, and 70? 

Analysis. In this operation 



2,5 


OPERATION. 

20 . . 12 . . 15 . 


.75 


2,3 


2.. 6. . 3 . 


. 15 


5 


5 



2x5x2x3x5 = 300, Ans. 



3,7 
2,5 



OPERATION. 

15 . . 42 . 



70 



2 ..10 



3x7x2x5 = 210, Ans. 



42 and 70 must contain 7 and 10, 

least common multiple of the remaining numbers, 15, 42, and 70 



we omit the 7 and 10, because 
they are exactly contained in 
some of the other given numbers ; 
thus, 7 is contained in 42, and 10 
in 70 ; and whatever will contain 
Hence we have only to find the 



LEAST COMMON MULTIPLE. 85 

From these examples we derive the following 

Rule. I. Write the numbers in a line, omitting such of the 
smaller numbers as are factors of the larger, and draw a vertical 
line at the left. 

II. Divide by any prime factor or factors that may be contained 
in one or more of the given numbers, and write the quotients and 
undivided numbers in a line underneath, omitting the Vs. 

III. In like manner divide the quotients and undivided numbers, 
and continue the process till all the factors of the given numbers 
have been transferred to the left of the vertical. Then multiply 
these factors together, and their product will be the hast common 
multiple required. 

Note. — We may use a composite number for a divisor, when it is contained 
in all the given numbers. 

EXAMPLES EOR PRACTICE. 

1. What is the least common multiple of 15, 18, 21, 24, 35, 
36, 42, 50, and 60 ? Ans. 12600. 

2. What is the least common multiple of 6, 8, 10, 15, 18, 20, 
and 24 ? Ans. 360. 

3. What is the least common multiple of 9, 15, 25, 35, 45, and 
100? An*. 6300. 

4. What is the least common multiple of 18, 27, 36, and 40 ? 

5. What is the least common multiple of 12, 26, and 52 ? 

6. What is the least common multiple of 32 ; 34, and 36 ? 

Ans. 4896. 

7. What is the least common multiple of 8, 12, 18, 24, 27, and 
36? 

8. What is the least common multiple of 22, 33, 44, 55, and 
.66? 

9. What is the least common multiple of 64, 84, 96, and 216 ? 

10. If A can build 14 rods of fence in a day, B 25 rods, C 8 
rods, and T> 20 rods, what is the least number of rods that will 
furnish a number of whole days' work to either one of the four 
men? Ans. 1400. 

8 



gg PROPERTIES OF NUMBERS. 

11. What is the smallest sum of money for which I can pur- 
chase either sheep at $4 per head, or cows at $21, or oxen at $49, 
or horses at $72 ? Arts. $3528. 

12. A can dig 4 rods of ditch in a day, B can dig 8 rods, and 
C can dig 6 rods; what must be the length of the shortest ditch, 
that will furnish exact days' labor either for each working alone 
or for all working together ? Arts. 72 rods. 

13. The forward wheel of a carriage was 11 feet in circumfer- 
ence, and the hind wheel 15 feet ; a rivet in the tire of each was 
up when the carriage started, and when it stopped the same rivets 
were up together for the 575th time ; how many miles had the 
carriage traveled, allowing 5280 feet to the mile ? 

Ans. 17 miles 5115 feet. 

CANCELLATION. 

157. Cancellation is the process of rejecting equal factors 
from numbers sustaining to each other the relation of dividend 
and divisor. 

158. It is evident that factors common to the dividend and 
divisor may be rejected without changing the quotient, (117, 
III). 

1. Divide 1365 by 105. 

Analysis. We first in- 

OPERATION. 

to^r ^ i. * -.o dicate the division by wri- 
1365 % x £ X t X 13 - ,. .. ,. ,; , t 
-ax _ — r . = 13 ting the dividend above a 

105 X £ X t horizontal line and the di- 

visor below. Then factor- 
ing each term, we find that 3, 5, and 7 are common factors ; and 
crossing, or canceling these factors, we have 13, the remaining factor 
of the dividend, for a quotient. 

159. If the product of several numbers is to be divided by 
the product of several other numbers, the common factors should 
be canceled before the multiplications are performed, for two 
reasons : 

1st. The operations in multiplication and division will thus be 
abridged. 



CANCELLATION. 87 

2d. The factors of small numbers are generally more readily 
detected than those of large numbers. 

2. Divide 20 times 56 by 7 times 15. 

operation. Analysis. Having first indi- 

4 g cated all the operations required 

^0 X ^ 32 by the question, we cancel 7 

~1 77 === 7~ =l"f from 7 and 56, and 5 from 15 

q and 20, leaving the factors 3 in 

the divisor, and 8 and 4 in the 
dividend. Then 8 X 4 = 32, which divided by 3, gives lOf , the quo- 
tient required. Hence the following 

Rule I. Write the numbers composing the dividend above a 
horizontal line, and the numbers composing the divisor below it. 

II. Cancel all the factors common to both dividend and divisor. 

III. Divide the product of the remaining factors of the dividend 
by the product of the remaining factors of the divisor, and the 
result will be the quotient 

Notes. — 1. When a factor is canceled, the unit, 1, is supposed to take its 
place. 

2. By many it is thought more convenient to write the factors of the dividend 
on the right of a vertical line, and the factors of the divisor on the left. 

EXAMPLES FOR PRACTICE. 

1. What is the quotient of 18 X 6 x 4 x 42 divided by 4 x 9 
x3x 7x6? 

FIRST OPERATION, SECOND OPERATION. 



*MxftxjLx M 



= 4 



*2 



fixfix $x tx$ g 

f 

4, Ans. 

2. Divide the product of 21 x 8 x 60 x 8 x 6 by 7 X 12 x 3 
X 8 x 3. Ans. 80. 

3. The product of the numbers 16, 5, 14, 40, 16, 60, and 50, 
m to be divided by the product of the numbers 40, 24, 50, 20, 7, 
and 10 ; what is the quotient ? Ans. 32. 



gg PROPERTIES OF NUMBERS. 

4. Divide the continued product of 12, 5, 183, 18, and 70 by 
the continued product of 3, 14, 9, 5, 20, and 6. 

5. If 213 x 84 x 190 x 264 be divided by 30 X 56 x 36, 
what will be the quotient ? 

6. Multiply 64 by 7 times 31 and divide the product by 8 
times 56, multiply this quotient by 15 times 88 and divide the 
product by 55, multiply this quotient by 13 and divide the pro- 
duct by 4 times 6. Ans. 403. 

7. How many cords of wood at $4 a cord, must be given for 3 
tons of hay at $12 a ton ? 

8. How many firkins of butter, each containing 56 pounds, at 
15 cents a pound, must be given for 8 barrels of sugar, each con- 
taining 195 pounds, at 7 cents a pound ? Ans. 13. 

9 . A grocer sold 16 boxes of soap, each containing 66 pounds 
at 9 cents a pound, and received as pay 99 barrels of potatoes, 
each containing 3 bushels ; how much were the potatoes worth a 
bushel ? 

10. A farmer exchanged 480 bushels of corn worth 70 cents a 
bushel, for an equal number of bushels of barley worth 84 cents a 
bushel, and oats worth 56 cents a bushel ; how many bushels of 
each did he receive ? Ans. 240. 

11. A merchant sold to a farmer two kinds of cloth, one kind at 
75 cents a yard, and the other at 90 cents, selling him twice as 
many yards of the first kind as of the second. He received as pay 
132 pounds of butter at 20 cents a pound ; how many yards of 
each kind of cloth did he buy ? 

Ans. 22 yards of the first, and 11 yards of the second. 

12. A man took six loads of potatoes to market, each load con- 
taining 20 bags, and each bag 2 bushels. He sold them at 44 
cents a bushel, and received in payment 8 chests of tea, each con- 
taining 22 pounds ; how much was the tea worth a pound ? 

Ans. 60 cents. 



DEFINITIONS, NOTATION, AND NUMERATION. 89 

FRACTIONS. 

DEFINITIONS, NOTATION, AND NUMERATION. 

160. When it is necessary to express a quantity less than a 
unit, we may regard the unit as divided into some number of equal 
parts, and use one of these parts as a new unit of less value than 
the unit divided. Thus, if a yard, considered as an integral unit, 
be divided into 4 equal parts, then one, two, or three of these 
parts will constitute a number less than a unit. The parts of a 
unit thus used are called fractional units ; and the numbers formed 
from them, fractional numbers. Hence 

161* A Fractional unit is one of the equal parts of an inte- 
gral unit. 

163. A Fraction is a fractional unit, or a collection of frac- 
tional units. 

163. Fractional units take their name, and their value, from 
the number of parts into which the integral unit is divided. Thus, 

If a unit be divided into 2 equal parts, one of the parts is 
called one half If a unit be divided into 3 equal parts, one of the 
parts is called one third. If a unit be divided into 4 equal parts, 
one of the parts is called one fourth. 

And it is evident that one third is less in value than one half 
one fourth less than one third, and so on. 

164:. To express a fraction by figures, two integers are re- 
quired ; one to denote the number of parts into which the inte- 
gral unit is divided, the other to denote the number of parts taken, 
or the number of fractional units in the collection. The former 
is written below a horizontal line, the latter above. Thus, 



One half is written «| 

One third " \ 

Two thirds " § 

One fourth " \ 



Two fourths " f 

hree 
8* 



Three fourths " | 



One fifth is written \ 
Two fifths " | 

One seventh " ^ 



Three eighths " | 

Five ninths " | 

Eight tenths " ft 



90 FRACTIONS. 

IGo, The Denominator of a fraction is the number below the 
line. 

It denominates or names the fractional unit, and it shows how 
many fractional units are equal to an integral unit. 

166. The Numerator is the number above the line. 

It numerates or numbers the fractional units; and it shows 
how many are taken. 

167. The Terms of a fraction are the numerator and deno- 
minator, taken together. 

168. Since the denominator of a fraction shows how many 
fractional units in the numerator are equal to 1 integral unit, it 
follows, 

I. That the value of a fraction in integral units, is the quo- 
tient of the numerator divided by the denominator. 

II. That fractions indicate division, the numerator being a 
dividend and the denominator a divisor. 

160. To analyze a fraction is to designate and describe its 
numerator and denominator. Thus f is analyzed as follows : — 

7 is the denominator, and shows that the units expressed by the 
numerator are sevenths, 

5 is the numerator, and shows that 5 sevenths are taken. 

5 and 7 are the terms of the fraction considered as an expres- 
sion of division, 5 being the dividend and 7 the divisor. 

EXAMPLES FOR PRACTICE. 

Express the following fractions by figures : — 

1. Four ninths. Ans. $. 

2. Seven fifty -sixths. Ans* £$. 

3. Sixteen forty-eighths. 

4. Ninety-five one hundred seventy -ninths. 

5. Five hundred thirty-six four hundredths. 

6. One thousand eight hundred fifty-seven nine thousand five 
hundred twenty-firsts. 

7. Twenty-five thousand eighty -sevenths. 

8. Thirty ten thousand eighty-seconds. 

9. One hundred one ten millionths. 



DEFINITIONS, NOTATION, AND NUMERATION. 91 



Read and analyze the following fractions : — 

11 4 . _7_ . 17 . 4_5_ . _72_ . _4 8 . _8 4_ . 4 5 6 
X - L * 9> 12> 38 > iOO> 375> 100^^ 7863J 53T* 

19 3_0 . 8 7. 9 5_ . 48. 75 . 175. 4 3_6 . _7 8J5 
• L - J * 4 J 30 J TOO J r2">".4-8 7^ 2 > 50 > 487$' 

1Q 467. 536. 10_0 00. __7_5 . 5 07 

XO ' 936 > 248 J 73 J 10000> 3007* 

14. 150. 436. 13785. 15 0072. 100001 
■ L ^ # 537> 972^ 47956~J 475000> 200002* 

Fractions are distinguished as Proper and Improper. 

17©. A Proper Fraction is one whose numerator is less than 
its denominator; its value is less than the unit 1. 

171. An Improper Fraction is one whose numerator equals 
or exceeds its denominator; its value is never less than the unit 1. 

Notes. — 1. The value of a proper fraction, always being less than a unit, can 
only be expressed in a fractional form ; hence, its name. 

2. The value of an improper fraction, always being equal to, or greater than 
a unit, can always be expressed in some other form; hence its name. 

173. A Mixed Number is a number expressed by an integer 
and a fraction. 

173* Since fractions indicate division, (168, II), all changes 
in the terms of a fraction will affect the value of that fraction ac- 
cording to the laws of division; and we have only to modify the 
language of the General Principles of Division, by substituting 
the words numerator, denominator, and fraction, or value of the 
fraction, for the words dividend, divisor, and quotient, respectively, 
and we shall have the following 

GENERAL PRINCIPLES OP FRACTIONS. 

174. Prin. I. Multiplying the numerator multiplies the 
fraction, and dividing the numerator divides the fraction. 

Prin. II. Multiplying the denominator divides the fraction, 
and dividing the denominator multiplies the fraction. 

Prin. III. Multiplying or dividing both terms of the fraction 
by the same number, does not alter the value of the fraction. 

17 O. These three principles may be embraced in one 

GENERAL LAW. 

A change in the NUMERATOR produces a LIKE change in the 
value of the fraction ; but a change in the DENOMINATOR produces 
an OPPOSITE change in the value of the fraction. 



92 FRACTIONS. 

REDUCTION. 
1 7G. The Reduction of a fraction is the process of changing 
its terms, or its form, without altering its value. 

CASE I. 

177. To reduce fractions to their lowest terms. 

A fraction is in its lowest terms when its numerator and denomi- 
nator are prime to each other ; that is, when both terms have no 
common divisor. 

I. Reduce the fraction j 6 q ^ to its lowest terms. 

operation. Analysis. Dividing both terms of 

_6_o__ = 12 _. 4 the fraction by the same number does 

Q r not alter the value of the fraction, 

15V 6_o _L 4 (174, III) ; hence, we divide both 

)l °^ " 7 terms of T %% by 5, and both terms of 

the result, £f , by 5, and obtain 4 for the final result. As 4 and 7 are 

prime to each other, the lowest terms of f£ 5 are %. 

Instead of dividing by the factors 5 and 3 successively, we may 
divide by the greatest common divisor of the given terms, and reduce 
the fraction to its lowest terms at a single operation. Hence, the 

Rule. Cancel or reject all factors common to both numerator 
and denominator. Or, 

Divide both terms by their greatest common divisor. 

EXAMPLES FOR PRACTICE. 

1. Reduce ^^ to its lowest terms. Ans. § 

2. Reduce || to its lowest terms. Ans. |. 

3. Reduce ||| to its lowest terms. Ans. f . 

4. Reduce -j?^ to its lowest terms. Ans. f . 

5. Reduce |§§ to its lowest terms. Ans. |. 

6. Reduce §§§ to its lowest terms. 

7. Reduce ||| to its lowest terms. 

8. Reduce j 6 ^ 9 ^ to its lowest terms. 

9. Reduce f §|| to its lowest terms. 

10. Reduce |4| to its lowest terms. Ans. -if* 

Note. — Consult the factor table. 

II. Reduce j^g to its lowest terms. Am. ||. 



REDUCTION. 93 

12. Reduce |i|| to its lowest terms. Ans. §f. 

13. Reduce |||f to its lowest terms. 

14. Reduce ffjf to its lowest terms. Ans. | {. 

15. Reduce |||||, and ||^f | to their lowest terms. 

CASE II. 

178. To reduce an improper fraction to a whole or 
mixed number. 

1. Reduce 2 T 9 / to a wn °l e or mixed number. 

OPERATION. 

yg = 297 ~- 12 = 24 T % = 24f . 

Analysis. Since the value of a fraction in integral units is equal 
to the quotient of the numerator divided by the denominator, (168, 1,) 
we divide the given numerator, 297, by the given denominator, 12, 
and obtain for the value of the fraction, the mixed number 24 T 9 g = 24f . 
Hence the 

Rule. Divide the numerator hy the denominator. 

Notes. 1. When the denominator is an exact divisor of the numerator, the 
result will be a whole number. 

2. In all answers containing fractions, reduce the fractions to their lowest 
terms. 

EXAMPLES FOR PRACTICE. 

1. Reduce 9 ^ 6 to an equivalent integer. Ans. 16. 

2. Reduce \ 6 to an equivalent integer. 

3. Reduce 1 § 4 to a mixed number. Ans. 17 *j. 

4. Reduce s j 4 ^ to a mixed number. Ans. 26-l-i. 

5. Reduce 5 5 ^ 8 to a mixed number, Ans. 24|. 

6. Reduce 9 ^ to a mixed number. Ans. 17*>§. 

7. Reduce 2 |f * to a mixed number. 

8. Reduce 3 ff| 5 to a mixed number. Ans. 156g. 

9. Reduce 3 f § 2 to a mixed number. 

10. Reduce 3 ^$ to a mixed number. Ans. 4^-J. 

11. Reduce 2 §f§ 5 to a mixed number. Ans. lOOf . 

12. Reduce j|}f $f to a mixed number. 

13. In 78 g 59 of a day how many (Jays? Ans. 982f days. 

14. In 4 T °3 7 of a dollar how many dollars? Ans. $31^. 

15. If 1000 dollars be distributed equally among 36 men, what 
part of a dollar must each man receive in change ? Ans. |. 



94 



FRACTIONS. 



CASE III. 

179. To reduce a whole number to a fraction having 
a given denominator. 

1. Reduce 37 to an equivalent fraction whose denominator shall 

be 5. 

operation. Analysis. Since in each unit there are 

37 X 5 = 185 5 fifths, in 37 units there must be 37 times 

37 — l 85 ? jl ns ^ 5 fifths, or 185 fifths = '§ 5 . The nume- 

rator, 185, is obtained in the operation by 
multiplying the whole number, 37, by the given denominator, 5. 
Hence the 

Rule. Multiply the whole number by the given denominator ; 
take the product for a numerator, under which write the given de- 
nominator. 

Note. — A whole number may be reduced to a fractional form by writing 1 
under it for a denominator; thus, 9 = £. 

EXAMPLES FOR PRACTICE. 

1. Reduce 17 to an equivalent fraction whose denominator 
shall be 6. Arts. 1 g 2 . 

2. Change 375 to a fraction whose denominator shall be 8. 

3. Change 478 to a fraction whose denominator shall be 24. 

4. Reduce 36 pounds to ninths of a pound. 

5. Reduce 359 days to sevenths of a day. Ans. 25 ^ 13 . 

6. Reduce 763 feet to fourteenths of a foot. Ans. 1 ° T 6 / 2 . 

7. Reduce 937 to a fractional form. Ans. 9 | 7 . 

CASE IV. 

ISO. To reduce a mixed number to an improper frac- 
tion. 
1. In 12f how many sevenths? 

operation. Analysis. In the whole number 12, there a^e 

12f 12 X 7 sevenths = 84 sevenths, (Case III), arid 

J_ 84 sevenths -f- 5 sevenths = 89 sevenths, or \ 9 . 

89 Hence the following 



REDUCTION. 95 

Rule. Multiply the whole number by the denominator of the 
fraction ; to the product add the numerator, and under the sum 
write the denominator. 

EXAMPLES FOR PRACTICE. 

1. Reduce 15f to fifths. Ans. \ 9 . 

2. Reduce 24| to an improper fraction. Ans. ^ 9 . 

3. Reduce 57f to an improper fraction. 

4. Reduce 356{f to an improper fraction. Ans. 6 ^| 4 . 

5. Reduce 872 T 5 2 to an improper fraction. 

6. Reduce 300 3 i^ to an improper fraction. Ans. 9 |^ 1 . 

7. Reduce 434|f to an improper fraction. Ans. 10 5 °3 00 . 

8. In 15| how many eighths ? 

9. In 135^ 3 q how many twentieths? Ans. 2 |§ a . 

10. In 43| bushels how many fourths of a bushel ? 

11. In 760 t 9 q days how many tenths of a day ? 

case v. 

181. To reduce a fraction to a given denominator. 

We have seen that fractions may be reduced to lower terms by 
division. Conversely, 

I. Fractions may be reduced to higher terms by multiplication. 

II. All higher terms of a fraction must be multiples of its 
lowest terms. 

1. Reduce | to a fraction whose denominator is 40. 

operation. Analysis. "We first divide 40, the re- 

40"-?- 8 = 5 quired denominator, by 8, the denomi- 

3^5 nator of the given fraction, to ascertain 

g x 5 = H> Ans if it be a multiple of this term, 8. The 
division shows that it is a multiple, and 
that 5 is the factor which must be employed to produce it. We there- 
fore multiply both terms of § by 5, (174, III), and obtain £§ , the re- 
quired result. Hence the 

Rule, Divide the required denominator by the denominator 
of the given fraction, and multiply both terms of the fraction by 
the quotient. 



96 FRACTIONS. 

EXAMPLES FOR PRACTICE. 

1. Reduce § to a fraction having 24 for a denominator, 
to a fraction whose denominator is 96 



Ans. || . 



T2 



Ans. §§. 



3. Reduce {§ to a fraction whose denominator is 51. 

4. Reduce T 9 3 to a fraction whose denominator is 78. 

5. Reduce g 6 ^ to a fraction whose denominator is 3000. 



An* 496 
J±nb ' 3000* 

6. Change 7f to a fraction whose denominator is 8. 

7. Change 16^ to a fraction whose denominator is 176. 

8. Change 5 T 3 T to a fraction whose denominator is 363. 

9. Change 36f to a fraction whose denominator is 42. 

Ans. if! 2 . 

CASE VI. 

183. To reduce two or more fractions to a common 
denominator. 

A Common Denominator is a denominator common to two or 
inore fractions. 

1. Reduce | and | to a common denominator. 

Analysis. We multiply the terms of the 
operation. fi rs t f rac tion by the denominator of the second, 

3 X 9 == 27 and the terms of the second fraction by the 

5x9 denominator of the first, (174, III). This 

7 w 5 must reduce each fraction to the same deno- 

q~ r "45 minator, for each new denominator will be the 

product of the given denominators. Hence the 

Rule. Multiply the terms of each fraction by the denominators 
of all the other fractions. 

Note. — Mixed numbers must first be reduced to improper fractions. 

EXAMPLES FOR PRACTICE. 

1. Reduce § and | to a common denominator. Ans. jf, T %. 

2. Reduce § and § to a common denominator. Ans. Jf , ||. 



REDUCTION. 97 

3. Reduce §, T 5 2 and | to a common denominator. 

An* 72 50 60 

juis. T5IJ , T2 q, T2]5 . 

4. Reduce i, 5| and 1| to equivalent fractions having a com- 
mon denominator. Ans. |§, y^ , ||. 

5. Reduce T 4 ^ and T 3 T to a common denominator. 

Ans - M> 2t¥t- 

6. Reduce ^, \ and y^ to a common denominator. 

7. Reduce |, |, T ^ and T 9 g to a common denominator. 

J OT <. 768 1152 896 864 
CASE VII. 

183. To reduce fractions to their least common de- 
nominator. 

The Least Common Denominator of two or more fractions is 
the least denominator to which they can all be reduced. 

184. We have seen that all higher terms of a fraction must 
be multiples of its lowest terms, (181, II). Hence, 

I. If two or more fractions be reduced to a common denomi- 
nator, this common denominator will be a common multiple of the 
several denominators. 

II. The least common denominator must therefore be the least 
common multiple of the several denominators. 

1. Reduce §, ^ and f^ to their least common denominator. 

operation. Analysis. We first find the least 



3 ,5 

2 , 2 



12 . . 15 common multiple of the given deno- 

7 minators, which is 60. This must 

be the least common denominator to 



3x5x2x2 = 60 which the given fractions can be re- 

duced, (II). Reducing each frac- 
a tion to the denominator 60, by Case 

V, we obtain |g, \\ and ^ for the 
answer. Hence the following 

Rule. I. Find the least common multiple of the given denom- 
inators, for the least common denominator. 
9 G 



5 

6 ~ 


50 
£0 


7 

T2" 


35 

6~0 


2 

1 5 — 


g 8 o 



98 FRACTIONS. 

II. Divide this common denominator by each of the given de- 
nominators, and multiply each numerator by the corresponding 
quotient. The products will be the new numerators. 

Notes. — 1. If the several fractions are not in their lowest terms, they should 
be reduced to their lowest terms before applying the rule. 

2. When two or more fractions are reduced to their least common denominator, 
their numerators will be prime to each other. 

EXAMPLES FOR PRACTICE. 

1. Keduce f and t 3 q to their least common denominator. 

Ano 25 12 

2. Reduce f , f and § to their least common denominator. 

3. Reduce f , T 7 5 and j^ to their least common denominator. 

4. Reduce f , f and | to their least common denominator. 

5. Reduce r 6 ? , %% and || to their least common denominator. 

An* 36 35 26 
JMS. 84? 84? H4' 

6. Reduce f , T ^, §§ and -^ to their least common denominator. 

An* 52 24 75 8 
jlus. 78 -, 7 g, 1B , ^g. 

7. Reduce 2|, T 7 g, ^ and §§ to their least common denomi- 
nator Am* 3 12 56 25 74 
naior. jltis. T2U , T2 -q, jsq, T3Z ^. 

8. Reduce |^, 5 9 g and || to their least common denominator. 

9. Reduce §§, T 2 ^% and ^| to their least common denominator. 

An? 60 20 21 

10. Reduce 5 ^, T 7 T 7 ^ and ||| to their least common denomi- 
nator An* 34 143 60 
naior. Jins. 25T , 52T , 2 2T . 

11. Reduce £f £, ||f and {^|j to their least common denomi- 
nator An* 371 901 713 

naior. Jins. j^j-g, t~zi^> T2~t^* 

12. Reduce 2^, j\ and 1 T 7 ^ to their least common denominator. 

13. Reduce j^g, §4|| and fgf § to their least common denom- 
inator An* 180614 193874 323733 

maior. jins. 3^4^55, 354^3^? 3 54F2-6* 

14. Reduce 4> jl? t 2 5> 57? 33 an( * I? to tne * r * east common 
denominator Ann 5400 6930 1008 2240 19 4 4 3213 

UtJIlUlUlUdtur. J±ns. 775^9, 7T6~0> T^SU' ^IgO> 7 5 SO? 75^6 0' 

15. Reduce %, r ^, yj> B , ^ and r y^ to their least common de- 
nominator. 

16. Reduce T ^, 7 \, §§ and 4^ to their least common denomi- 
nator Am* 28 7 60 455 

UdTOr - Ans. Jjys , TTJ ^, rTJ ^, TTr5 . 



ADDITION. 99 

ADDITION. 

185. The denominator of a fraction determines the value of 
the fractional unit, (1G5) ; hence, 

I. If two or more fractions have the same denominator, their 
numerators express fractional units of the same value. 

II. If two or more fractions have different denominators, their 
numerators express fractional units of different values. 

And since units of the same value only can be united into one 
sum, it follows, 

III. That fractions can be added only when they have a com- 
mon denominator. 

1. What is the sum of ±, T 5 2 and f^t 

OPERATION. 



_ 12 + 25 + 8 _ 
60 



1 J_ _5_ I _2 ' * 4 5 3 



J_ 5 I 2 

i 12 ^ IS — 



Analysis. We first reduce the given fractions to a common deno- 
minator, (III). And as the resulting fractions, Jf , fff, and ■£$ have the 
same fractional unit, (I), we add them by uniting their numerators 
into one sum, making £| === §, the answer. 

2. Add 5f , 3| and 4 T 7 3 . 



Analysis. The sum of the 

integers, 5, 3, and 4, is 12; the 

3 i 

TTonr»A +.Vi 
1 2' xo ^24 



OPERATION. 

5 + ? + i = 12 sum of the fractions, f, J, and 

4 ' 8 "t" T2 ~~_~2? -X, is 2/3 . Hence, the sum of 

14^, Arts. both fractions and integers is 
12+2^ = 14^. 

186, From these principles and illustrations we derive the 
following general 

Rule. I. To add fractions When necessary, reduce ihz frac- 
tions to their least common denominator ; then add the numerators 
and place the sum over the common denominator. 

II. To add mixed numbers. — Add the integers and fractions 
separately, and then add their sums. 

Note. — All fractional results should be reduced to their lowest terms, and if 
improper fractions, to whole or mixed numbers. 



100 



FRACTIONS. 



EXAMPLES FOR PRACTICE. 



1. What is the sum of j\, T 4 3 , T s 2 and ±| ? Am. 2f 

2. What is the sum of if, T %, T \ and ^ ? .4ms. If. 

3. What is the sum of 2 5 T , £, ^f and ^f ? 

4. What is the sum of 1\\, 8§§, 21}, 5|| and 4|| ? 



u4«s. 28|. 



5. 


What is the sum of 37g 9 g , 


6. 


Add |, | and §. 


7. 


Add |, |, if and T V 


8. 


Add I, | and T L. 


9. 


Add r %, if and &. 


10. 


A AA 13 11 aTlf ] 29 


11. 


AddI,H,i|,||and|f 


12. 


Add 5|, 4| and 2ft. 


13. 


Add 16 J 2 and 24 J g . 


14. 


Add 4 2f , 3|, 4| and 5| 


15. 


Add 4ft, 8 2 \ and 2ft. 


16. 


Add l> h T2 and T7- 


17. 


Add |, |, ft and T \. 


18. 


Add |, |, ft, ft and ||. 


19. 


Add i, ft, ft and |. 



12f|,13fgandf§? 



-In*. 23V 



Ans. llf 
jins. 2||. 
Ans 4 T %. 



.4ns. 40^. 



-4rcs. 14lf . 
37 

5T* 



4flt. Iff. 



20. Add 411, 105§, 300|, 241| and 472|. Ans. 1161§9. 

21. Add 4|, 21, 1 T ^, 2&, 5ft, 7f, 4£ and 6|. 

22. Four cheeses weighed respectively 36§, 42|, 39 T 7 g and 511 
pounds ; what was their entire weight ? Ans. 169|| pounds. 

23. What number is that from which if 4| be taken, the re- 
mainder will be 3||? Ans. 8f, 

24. What fraction is that which exceeds T 5 H by ^ ? 

25. A beggar obtained l of a dollar from one person, l from 
another, l from another, and T \ from another ; how much did he 
get from all ? 

26. A merchant sold 46| yards of cloth for $127/^, 64^ yards 
for $226§, and 76| yards for $312§ ; how many yards of cloth did 
he sell, and how much did he receive for the whole ? 

Ans. 187|| yards, for $666}§. 



SUBTRACTION. 101 

SUBTRACTION. 

18 7. The process of subtracting one fraction from another is 
based upon the following principles : 

I. One number can be subtracted from another only when the 
two numbers have the same unit value. Hence, 

II. In subtraction of fractions, the minuend and subtrahend 
must have a common denominator. 

1. From | subtract f . 

operation. Analysis. Reducing the 

4 2 _ 1 2j^_i o __. _2_ given fractions to a common 

denominator, the resulting 
fractions j| and }■§ express fractional units of the same value, (185, 
I). Then 12 fifteenths less 10 fifteenths equals 2 fifteenths = T 2 3 , the 
answer. 

2. From 2381 take 24|. 

operation. Analysis. We first reduce the frac- 

238^ = 238-% tional parts, £ and f, to the common 

245 __. 24!£ denominator, 12. Since we cannot 

take jf from T 3 2, we add 1 = jf, to T 3 2, 

213 T 5 2 Ans. making f§. Then, j| subtracted from 
If leaves ^ ; and carrying 1 to 24, the integral part of the subtrahend, 
(73, I), and subtracting, we have 213 T 5 2 for the entire remainder. 

188. From these principles and illustrations we derive the 
following general 

Kule. I. To subtract fractions. — When necessary, reduce the 
fractions to their least common denominator. Subtract the nume- 
rator of the subtrahend from the numerator of the minuend, and 
place the difference of the new numerators over the common denom- 
inator. 

II. To subtract mixed numbers. — Reduce the fractional parts 

to a common denominator, and then subtract the fractional and 

integral parts separately. 

Note. — We may reduce mixed numbers to improper fractions, and subtract 
1 by the rule for fractions. But this method generally imposes the useless labor 
of reducing integral numbers to fractions, and fractions to integers again. 

9* 



102 



J2JXJ\K;Xl\J±XtD. 

EXAMPLES FOR PRACTICE. 




From A take T 3 ^. 


Ans. A. 


From || take ||. 


J.WS. |. 


From || take -fy. 




From | take f . 


4«w. |f 


From | take ^. 




From If take <£. 


^S. f j. 


From T 9 4 take J$. 


j4?w. ^ g . 


From l| take §f . 


^is. T V 


From / T take T x ^. 




From T 7 ^ take ? %. 


-4m. ||. 


From & take T ^. 


Ans. sI-q. 


From /g take $. 




From 16| take 7£. 


Ans. 9f . 


From 361| take 8£f . 


^trcs. 27|. 


From 25 T 7 n subtract 14{§. 




From 75 subtract 4f . 


.4ns. 704. 


From 18| subtract 5|. 




From 26/ ? subtract 25{§. 




From 28 J| subtract 3^. 


Jite. 24j|. 



1. 

2. 

3. 

4. 

5 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 



16. 
17. 

18. 

19. 

20. From 78/5 subtract 32§. 
' 21. The sum of two numbers is 26|, and the less is 7 T ^ ; what 
is the greater ? Ans. 19/ 3 . 

22. What number is that to which if you add 18|, the sum 
willbe97f? 

23. What number must you add to the sum of 126^ and 240|, 
to make 560f ? Am. 193||. 

24. What number is that which, added to the sum of ^, T 5 5 , 
and Jg, will make §§ ? Ans. |^. 

25. To what fraction must § be added, that the sum may be f ? 
26 From a barrel of vinegar containing 31} gallons, 14i gallons 

were drawn ; how much was then left ? Ans. 16| gallons. 

27. Bought a quantity of coal for $140f, and of lumber for 
$456f . Sold the coal for $775J, and the lumber for $516 T 3 g ; how 
much was my whole gain ? Ans. $694||. 






THEORY OF MULTIPLICATION AND DIVISION 1Q3 



THEORY OF MULTIPLICATION AND DIVISION OF FRACTIONS. 

189. In multiplication and division of fractions, the various 
operations may be considered in two classes : 

1st. Multiplying or dividing a fraction. 
2d. Multiplying or dividing by a fraction. 

190. The methods of multiplying and dividing fractions may 
be derived directly from the General Principles of Fractions, 
(174); as follows: 

I. To multiply a fraction. — Multiply its numerator or divide its 
denominator j (1741, I. and IT). 

II. To divide a fraction. — Divide its numerator or multiply its 
denominator, (174, I. and II). 

GENERAL LAW. 

III. Perform the required operation upon the numerator, or the 
opposite upon the denominator, (174, III). 

191. The methods of multiplying and dividing by a fraction 
may be deduced as follows : 

1st. The value of a fraction is the quotient of the numerator 
divided by the denominator (168, I). Hence, 

2d. The numerator alone is as many times the value of the 
fraction, as there are units in the denominator. 

3d. If, therefore, in multiplying by a fraction, we multiply by 
the numerator, this result will be too great, and must be divided 
by the denominator. 

4th. But if in dividing by a fraction, we divide by the nume- 
rator, the resulting quotient will be too small, and must be multi- 
plied by the denominator. 

Hence, the methods of multiplying and dividing by a fraction 
may be stated as follows : 

I. To multiply by a fraction. — Multiply by the numerator and 
divide by the denominator, (3d). 

II. To divide by a fraction. — Divide by the numerator and mulr 
tivly by the denominator, (4th). 



104 



FRACTIONS. 



GENERAL LAW. 
III. Perform the required operation by the numerator and the 
opposite by the denominator. 



MULTIPLICATION. 



193. 1. 



Multiply T 5 2 by 4. 



Analysis. In the first opera- 
tion, we multiply the fraction 
by 4 by multiplying its nume- 
rator by 4 ; and in the second 
operation, we multiply the frac- 
tion by 4 by dividing its denom- 
inator by 4, (190, I or III). 

In the third operation, we ex- 
press the multiplier in the form 
of a fraction, indicate the mul- 
tiplication, and obtain the result by cancellation. 

2. Multiply 21 by f 



FIRST 


OPERATI 


ON. 


5 


X4: 


— 12 


-1^ 
X 3 


SECOND 


OPERATION. 


5 


X4 


5 _ 

— 3 


: 1^ 
x 3 


THIRD 


OPERATION. 


5 

A2 


4 


=== 3 == 


If 


3 









FIRST OPERATION. 

21 X 



4 84 12 



SECOND 

21x4 = 3x 



OPERATION. 

4 = 12 



Analysis. To multiply by 4> 
we must multiply by 4 and di- 
vide by 7, (191, I or III). 

In the first operation, we first 
multiply 21 by 4, and then di- 
vide the product, 84, by 7. 

In the second operation, we 
first divide 21 by 7, and then 
multiply the quotient, 3, by 4. 

In the third operation, we ex- 
press the whole number, 21, in 
the form of a fraction, indicate the multiplication, and obtain the 
result by cancellation. 



third operation. 



3. Multiply T % by g. 

first operation. 

1st step, 
2d step, 



5 
11 


V 7 — 3 5 
A < — 14 


35 


•- 8 = - 3 -4r 




35 5 

TT2 — TS 



= A Ans. 



ANALYSfs. To multiply by 
J, we must multiply by 7 and 
divide by 8, (191, I or III). 
In the first operation, we mul- 
tiply T 5 ^ by 7 and obtain f f ; 



MULTIPLICATION. 105 

second operation. we then divide ?£ by 8 and obtain 

j 6 ! X | = T 3 T 5 2 = T 5 5 t 3 t 5 2> which reduced gives T 5 F , the 

required product. In the second 

THIRD OPERATION. , u . ,, i, 

operation we obtain the same result 

_ ^ _= J> by multiplying the numerators to- 

P-r* ° gether for the numerator of the pro- 

^ duct, and the denominators together 

for the denominator of the product. In the third operation, we indicate 

the multiplication, and obtain the result by cancellation. 

193. From these principles and illustrations we derive the 
following general 

Rule. I. Reduce all integers and mixed numbers to improper 

fractions. 

II. Multiply together the numerators for a new numerator ; and 

| the denominators for a new denominator. 

Notes. — 1. Cancel all factors common to numerators and denominators. 
2. If a fraction be multiplied by its denominator, the product will be the 
I numerator. 

EXAMPLES FOR PRACTICE. 

1. Multiply | by 8. Ans. 2f 

2. Multiply | by 27, T % by 4, and ft by 9. 



3. Multiply 7 \ by 15. Ans. §. 

4. Multiply 8 by f . Ans. 6. 

5. Multiply 75 by T % 7 by #, 756 by f , and 572 by $. 

6. Multiply | by §. 

7. Multiply 11 by ffy and if by § i. 

8. Multiply A by fi, and ^ by |i. 

9. Multiply 24 by 3§ . Ans. 6. 

10. Multiply If by iff. Ans. 2£. 

11. Multiply^ by 21|. 

12. Find the value of § X § X T 9 ¥ X g. t ^4ws. 2 \. 

13. Find the value of | X | X || X T 4 T X ff . J.ws. T V 

14. Find the value of Jf X ^ X iff. 

15. Find the value of 2| x 21 x T \ X T £ H X l/ ¥ X 26f 

J^s. 2. 

16. Find the value of T 7 T x 5 5 T X 4| x 15 x T 3 ff . 

17. Find the value of Jft x F fo * Vs - Am. ih- 



106 



FRACTIONS. 



18. Find the value of (4£ x |) + If X (3| — -&). 

19. Find the value of 28 + (7| — 2|) X | X (f + |> 

Note 3. — The word of between fractions is equivalent to the sign of multi- 
plication; and such an expression is sometimes called a compound fraction. 

Find the values of the following indicated products : — 

20. | of | of |. Am. §. 

21. | of | of 2 \. Ans. T \. 

22. | of & of f 

23. i^ of 3% of 3 f of ||. Ans. 5 %. 

24. £ of | of | of f of | of f of I of | of T %. 

In the following examples, cancellation may be employed by the 
aid of the Factor Table. 

25. What is the value of f £ \ x j||| X |||| ? -4ws. T 5 o 3 T . 

26. What is the value of fff \ X \\% \ X |||| ? Ans. £f 

27. What is the value of £§|f x ||f| X |f|f ? 

/J W o 1303 1 

28. What will 7 cords of wood cost at $3| per cord ? 

Ans. $25f. 

29. What is the value of (f) 2 X |f X (£) 5 ? ^s. ^f^. 

30. If 1 horse eat | of a bushel of oats in a day, how many 
bushels will 10 horses eat in 6 days ? Ans. 25f. 

31. What is the cube of 12| ? 

32. At $9| per ton, what will be the cost of £ of § of a ton of 
hay? Ans. $4. 

33. At $ T % a bushel, what will be the cost of If bushels of 
corn? 

34. When peaches are worth $| per basket, what is 4 of a 
basket worth ? 

35. A man owning | of 156| acres of land, sold ^ of f of his 
share; how many acres did he sell? Ans. 47. 

36. What is the product of (|) s X (|) 2 X ( T 9 <j) 2 X (3£)*? 

Ans. iff. 

37. If a family consume 1^ barrels of flour a month, how many 
barrels will 6 families consume in 8 t 9 q months? 

38. What is the product of 150£— 0} of 121§+f of 48|)— 75, 
multiplied by 3 x (| of 1| of 4— 2|) ? Ans. 266 ffl. 



DIVISION. 



107 



39. A man at his death left his wife $12,500, which was £ of 
| of his estate ; she at her death left f of her share to her 
daughter ; what part of the father's estate did the daughter re- 
ceive? Ans. || . 

40. A owned | of a cotton factory, and sold | of his share to 
B, who sold ^ of what he bought to C, who sold f of what he 
bought to D for $3000 \ what part of the whole did D own, and 
what was the value of the factory ? 

Ans. D owned ^ ; the factory was worth $14,400. 

41. What is the value of 2|x~J+ § of 4| x (|) 2 +(3|) 3 — 3|) 2 ? 

Ans. 36|f §. 
DIVISION. 



194. 1. Divide |i by 3. 



FIRST OPERATION. 



21 

25 



8 — ^ 

° — 25 



SECOND OPERATION. 

21_i_Q — 21 __> 7 
55 : ° ~ 75 ~ 25 

2. Divide 15 by f . 

FIRST OPERATION. 

15 -T- 1=5 x 7=35 

SECOND OPERATION. 

15 -T- 1 = 105 — 3 = 35 

In the second operation we first 
the product by 3. 

3. Divide T % by f. 

FIRST OPERATION. 



Analysis. In the first ope- 
ration we divide the fraction by 
3 by dividing its numerator by 
3, and in the second operation 
we divide the fraction by 3 by 
multiplying its denominator by 
3, (190, I or III). 

Analysis. To divide by f , we 
must divide by 3 and multiply 
by 7, (191, II or III). 

In the first operation, we first 
divide 15 by 3, and then mul- 
tiply the quotient by 7. 
multiply 15 by 7, and then divide 



1st step, -~z -f- 3 = 



1 5 
_4_ 
45 



_4 

45 



2d step, -fe X 5 = f | = | Ans, 



SECOND OPERATION. 



4 
IT* 



y 5 20 4 

X 3 — 4 5 — 3 



Analysis. To divide by 
|, we must divide by 3 and 
multiply by 5, (191, II or 
III). In the first operation 
we first divide T 4 y by 3 by 
multiplying the denomina- 
tor by 3. We then multi- 



108 



FRACTIONS. 



THIRD OPERATION. Pty tne result, ?3> by 5 > by 

4. k multiplying the numerator 

£X X 3 = I by 5, giving f £ = £ for the 

g required quotient. By in- 

specting this operation, we 
observe that the result, f §, is obtained by multiplying the denomi- 
nator of the given dividend by the numerator of the divisor, and the 
numerator of the dividend by the denominator of the divisor. Hence, 
in the second operation, we invert the terms of the divisor, |, and 
then multiply the upper terms together for a numerator, and the 
lower terms together for a denominator, and obtain the same result as 
in the first operation. In the third operation, we shorten the pro- 
cess by cancellation. 

We have learned (107) that the reciprocal of a number is 1 
divided by the number. If we divide 1 by |, we shall have 1 -=- 
| = 1 x | = f . Hence 

195. The Reciprocal of 9 Fraction is the fraction inverted. 

From these principles and illustrations we derive the following 
general 

Rule. I. Reduce integers and mixed numbers to improper 
fractions. 

II. Multiply the dividend by the reciprocal of the divisor. 

Notes. •— I. If the vertical line be used, the numerators of the dividend and 
the denominators of the divisor must be written on the right of the vertical. 

2. Since a compound fraction is an indicated product of several fractions, its 
reciprocal may be obtained by inverting each factor of the compound fraction. 

EXAMPLES FOR PRACTICE. 

1. Divide || by 4. if X | = &, Ans. 

2. Divide jf by 5, and iff by 80. 

3. Divide 10 by § . Ans. 35. 

4. Divide 28 by |. and 3 by &. 

5. Divide 56 by If. Ans. 36. 

6. Divide £f by §. 

7. Divide }J by f , '§ by f„ and 3| by 5f 

8. Divide If by If Ans. If 

9. Divide ltf by §§. Ans 



3' 
T3* 



DIVISION. 109 

10. Divide f of § by % of T \. 

operation. Analysis. The dividend, 

3^5 — i reduced to a simple fraction, 

7 v 5 5 is i; the divisor, reduced 

1? ^ 74 — TU 

1 ^ i_8 __. 6 __ J i ^ ns# in like manner, is T 5 ^ ; and 

Q r i divided by T % is li, the 

^ X 5 X * X U = 11 quotient required. Or, we 

5 ^ 7 5 " 5 ma y apply the general rule 
directly by inverting both factors of the divisor,. 

Note 3. — The second method of solution given above has two advantages. 
1st, It gives the answer by a single operation ; 2d, It affords greater facility for 
cancellation. 



11. 


Divide | of T \ by T \ of T %. 


^4tis. 1. 


12. 


Divide & of 7 % by f of 3 5 T . 


Arts. l|i 


13. 


Divide 2^ x 1\ by 3^ x 3-,%. 




14. 


Divide 11 by f X 5£ x 7. 




15. 


Divide 3^ x 19 by \ X 7f x If. 


.4?is. 25. 


16. 


Divide y % x if by J x J x A X ff 


V 5 1 

Ans. 3ff. 


17. 


Divide iu by iWh- 


Am. lsjfe 


18. 


Divide 2 %Wt by U X M X |f 


Ans. if. 


19. 


Divide ix|X|x|by|x|x| 


x S x TO" 


20. 


5J. 
Wbat is the value of ■— 1 
4 I 

OPERATION, 
*1 11 

J2— "22— 2" * .M.T" * — 

^5 5 






<ii 



Analysis. The fractional form indicates division, the numerator 
being the dividend and the denominator the divisor, (168, II) ; hence, 
we reduce the mixed numbers to improper fractions, and then treat 
the denominator, 2 /, as a divisor, and obtain the result, 1{, by the 
general rule for division of fractions. 

4| 



Note 4. — Expressions like -| and — are sometimes called complex fractions. 



*7 5 

5. In the reduction of complex fractions to simple fractions, if either the 
numerator or denominator consists of one or more parts connected by + or — , 
the operations indicated by these signs must first he performed, and afterward 
the division. 

3 

21. What is the value of |-? Ans. f. 

10 



HO FRACTIONS. 



2 v 1 1 



22. What is the value of \ * "" ? u4ns. 2. 



23. What is the value of ^^ ? Arts. 7A. 



3 3 



X T2 



24. Reduce y-r— | to its simplest form. 

3+5 
5 2 

25. Reduce ^ 1 to its simplest form. 

3 x 7 
5 Q f 3 

26. Reduce ^ — ^~- to its simplest form. Ans. 1{|. 



27. If 7 pounds of coffee cost $|, how much will 1 pound cost? 

28. If a boy earn $| a day, how many days will it take him to 
earn $6^? Arts. 17f 

29. If | of an acre of land sell for $30, what will an acre sell 
for at the same rate ? Ans. $67^. 

30. At | of | of a dollar a pint, how much wine can be bought 
for $ T % ? Ans. 2| pints. 

31. If T 3 ^ of a barrel of flour be worth $2|, how much is 1 
barrel worth I Ans. $7f . 

32. Bought \ of 41 cords of wood, for § of \ of $30; what 
was 1 cord worth at the same rate? Ans. $4 T ^. 

33. If 235! acres of land cost W725f , how much will 125j 
acres cost? Ans. $918|4f- 

34. Of what number is 26^ the § part? Ans. 31^. 

35. The product of two numbers is 27, and one of them is 2| ; 
what is the other ? 

36. By what number must you multiply 16{^ to produce 148| ? 

37. What number is that which, if multiplied by f of | of 2, 
will produce |? Ans. 4||. 

38. Divide 720 — (| X 28 — 7£) by 40| + ( T % ~ f ) x Q)*. 

39. What is the value of (s^ x (|) 2 + f of l) 8 ~ (l7£ — f 
+ f=7|) 8 x5)? 

40 Divide ^ of ® 8 x3i (fof5|) 22464 



2 



of (9i) 2 + (I)* " J T \ x 2* 



GREATEST COMMON DIVISOR. ;Q1 

GREATEST COMMON DIVISOR OP FRACTIONS. 

196. The Greatest Common Divisor of two or more fractions 

is the greatest number which will exactly divide each of them, 

giving a whole number for a quotient. 

Note. — The definition of an exact divisor, (128), is general, and applies to 
fractions as well as to integers. 

19 7. In the division of one fraction by another the quotient 
will be a whole number, if, when the divisor is inverted, the two 
lower terms may both be canceled. This will be the case when 
the numerator of the divisor is exactly contained in the numerator 
of the dividend, and the denominator of the divisor exactly 
contains, or is a multiple of, the denominator of the dividend. 
Hence, 

I. A fraction is an exact divisor of a given fraction when its 
numerator is a divisor of the given numerator, and its denominator 
is a multiple of the given denominator. And, 

II. A fraction is a common divisor of two or more given frac- 
tions when its numerator is* a common divisor of the given nume- 
rators, and its denominator is a common multiple of the given 
denominators. Therefore, 

III. The greatest common divisor of two or more given frac- 
tions is a fraction whose numerator is the greatest common divisor 
of the given numerators, and whose denominator is the least com- 
mon multiple of*the given denominators. 

1. What is the greatest common divisor of |, A, and {§? 

Analysis. The greatest common divisor of 5, 5, and 15, the given 
numerators, is 5. The least common multiple of 6, 12, and 16, the 
given denominators, is 48. Therefore the greatest common divisor 
of the given fractions is ^\, Ans. (HI). 
Proof. 



Prime to each other. 



108. From these principles and illustrations, we derive the 
following 



5 
6 


~ 4 8 — ° 


5 


_^_ 5 4. 

~ 48 ~ * 


T2 


15 


• 48 — V 



H2 FRACTIONS. 

Rule. Find the greatest common divisor of the given nume- 
rators for a new numerator, and the least common multiple of the 
given denominators for a new denominator. This fraction will be 
the greatest common divisor sought 

Note. — Whole and mixed numbers must first be reduced to improper fractions, 
and all fractions to their lowest terms. 

EXAMPLES FOR PRACTICE. 

1. What is the greatest common divisor of |, ^|, and ^f ? 

Ans. T f ^ 

2. What is the greatest common divisor of 3i, 1|, and ||? 

3. What is the greatest common divisor of 4, 2§, 2§, and -^ ? 

An* -ft. 

4. What is the greatest common divisor of 109^ and 122| ? 

5. What is the length of the longest measure that can be 
exactly contained in each of the two distances, 18| feet and. 57^ 
feet? Ans. 2 T % feet. 

6. A merchant has three kinds of wine, of the first 109| gal- 
lons, of the second 128| gallons, of the third 115| gallons ; he 
wishes to ship the same in the fewest casks of equal size that will 
contain each kind of wine. How many kegs will be required ? 

Ans. 59. 



LEAST COMMON MULTIPLE OF FRACTIONS. 

199. The Least Common Multiple of two or more fractions is 
the least number which can be exactly divided by each of them, 
giving a whole number for a quotient. 

200. Since in performing operations in division of fractions 
the divisor is inverted, it is evident that one fraction will exactly 
contain another when the numerator of the dividend exactly con- 
tains the numerator of the divisor, and the denominator of the 
dividend is exactly contained in the denominator of the divisor 
Hence, 

I. A fraction is a multiple of a given fraction when its nume- 
rator is a multiple of the given numerator, and its denominator is 
a divisor of the given denominator. And 



LEAST COMMON MULPIPLE. H3 

II. A fraction is a common multiple of two or more given frac- 
tions when its numerator is a common multiple of the given nume- 
rators, and its denominator is a common divisor of the given 
denominators. Therefore, 

III. The least common multiple of two or more given fractions 

is a fraction whose numerator is the least common multiple of the 

given numerators, and whose denominator is the greatest common 

divisor of the given denominators. 

Note. — The least whole number that will exactly contain two or more given 
fractions in their lowest terms, is the least common multiple of their numera- 
tors, (193, Note 2). 

1. What is the least common multiple of |, T 5 3? and ||? 

Analysis. The least common multiple of 3, 5, and 15, the given 
numerators, is 15 ; the greatest common divisor of 4, 12, and 16, the 
given denominators, is 4. Hence, the least common multiple of the 
given fractions is \ 5 = 3|, Ans. (III). 

201. From these principles and illustrations we derive the 
following 

Rule. Find the least common multiple of the given numerators 
for a new numerator, and the greatest common divisor of the given 
denominators for a new denominator. This fraction will be the 
least common multiple sought. 

Note. — Mixed numbers and integers should be reduced to improper fractions, 
and all fractions to their lowest terms, before applying the rule. 

EXAMPLES FOR PRACTICE. 

1. What is the least common multiple of |, T ^, j§, and -^ i 

Ans. 111. 

2. What is the least common multiple of ^, ||, and |$ ? 

3. What is the least common multiple of 2||, 1||, and ■££$ ? 

4. What is the least common multiple of £, f , f , f , §, f , g, f, 
and T %? Ans. 2520. 

5. The driving wheels of a locomotive are 15 T 5 g feet in circum- 
ference, and the trucks 9f feet in circumference. What distance 
must the train move, in order to bring the wheel and truck in the 
same relative positions as at starting ? Ans. 459| feet. 

10* H 



114 FRACTIONS. 



PROMISCUOUS EXAMPLES. 

1. Change | of f to an equivalent fraction having 135 for its 
denominator. Am. T ^ 5 g. 

2. Reduce |, i, g, and || to equivalent fractions, whose denom- 
inators shall be 48. 

3. Find the least common denominator of \\, f, 2, t 7 q, | of |, 
4 of l 

2 f 3 2_ f 5 

4. The sum of - — ^— ^ and / -j~ is equal to how many times 

their difference ? J.rcs. 2. 

543 15 

5. The less of two numbers is = — ~%~, and their difference -^ : 

what is the greater number ? ^4ws. 34 T 4 g 3 ^ 

6. What number multiplied by § of g X 3§, will produce §f ? 

-4ws. |. 

7. Find the value of £±3 X ^ + (2 + i) -~ (3 + 4) 
ll- 3 

, xx 32" J™- »7_33_ 

~r* ^glf'* -^ ns * *14U 

8. What number diminished by the difference between | and | 
of itself, leaves a remainder of 144 ? Am. 283^. 

9. A person spending |, §, and ^ of his money, had $119 left; 
how much had he at first ? 

10. What will \ of lOg cords of wood cost, at -fa of $42 per 
cord? Am. $31^. 

11. There are two numbers whose difference is 25 T 7 ^, and one 
number is f of the other ; what are the numbers ? 

Am. 63f and 89 T \. 

12. Divide $2000 between two persons, so that one shall have 
I as much as the other. Am. $1125 and $875. 

13. If a man travel 4 miles in § of an hour, how far would he 
travel in 1^ hours at the same rate ? Am. 10 miles. 

14. At $| a yard, how many yards of silk can be bought for 

15. How many bushels of oats worth $| a bushel, will pay for 
I of a barrel of flour at $7^ a barrel ? 



PROMISCUOUS EXAMPLES. ;Q5 

16. If | of a bushel of barley be worth f of a bushel of corn, 
and corn be worth $f per bushel, how many bushels of barley will 
$15 buy? Arts. 18. 

17. If 48 is f of some number, what is | of the same number ? 

18. If cloth 1| yards in breadth require 20^ yards in length to 
make a certain number of garments, how many yards in length 
will cloth | of a yard wide require to make the same ? 

19. A gentleman owning | of an iron foundery, sold | of his 
share for $2570§ ; how much was the whole foundery worth ? 

Ans. $5141f 

20. Suppose the cargo of a vessel to be worth $10,000, and | 
of f of y 9 ^ of the vessel be worth \ of f of If of the cargo ; what 
is the whole value of the ship and cargo? Ans. $18333^. 

21. A gentleman divided his estate among his three sons as fol- 
lows : to the first he gave f of it ; to the second f of the remain- 
der. The difference between the portions of first and second was 
$500. What was the whole estate, and how much was the third 
son's share ? A f Whole estate, $12000. 

I Third son's share, $2500. 

22. If 7| tons of hay cost $60, how many tons can be bought 
for $78, at the same rate ? 

23. If a person agree to do a job of work in 30 days, what part 
of it ought he to do in 16| days ? Ans. ±±. 

24. A father divided a piece of land among his three sons ; to 
the first he gave 12| acres, to the second f of the whole, and to 
the third as much as to the other two ; how many acres did the 
third have? Ans. 49 acres. 

25. If | of 6 bushels of wheat cost $4^, how much will f of 1 
bushel cost? 

26. A man engaging in trade lost § of his money invested, after 
which he gained $740, when he had $3500 ; how much did he 
lose? Ans. $1840. 

27. A cistern being full of water sprung a leak, and before it 

could be stopped, f of the water ran out, but | as much ran in at 

the same time ; what part of the cistern was emptied ? 

i 

4' 



216 FRACTIONS. 

28. A can do a certain piece of work in 8 days, and B can do 
the same in G days; in what time can both together do it? 

Ans. 3| days. 

29. A merchant sold 5 barrels of flour for $32 £, which was £ 
as much as he received for all he had left, at $4 a barrel ; how 
many barrels in all did he sell ? Ans. 18. 

30. What is the least number of gallons of wine, expressed by 
a whole number, that can be put without waste into bottles con- 
taining |, |, f, and | gallons each? Ans. 60. 

31. A, B, and C start at the same point in the circumference 
of a circular island, and travel round it in the same direction. A 
makes § of a revolution in a day, B T 4 7 , and C B 8 T . In how many 
days will they all be together at the point of starting ? 

Ans. 178 J days. 

32. Two men are 64f miles apart, and travel toward each other; 
when they meet one has traveled 5 J miles more than the other; 
how far has each traveled ? 

Ans. One 29| miles, the other 35i miles. 

33. There are two numbers whose sum is l T \j, and whose dif- 
ference is |; what are the numbers? Ans. | and ^. 

34. A, B, and C own a ferry boat ; A owns y 1 ^ of the boat, 
and B owns T 7 H of the boat more than C. What shares do B and 
C own respectively? Ans. B, T 9 ^ ; C, £f. 

35. A schoolboy being asked how many dollars he had, replied, 
that if his money be multiplied by j|, and £ of a dollar be added 
to the product, and | of a dollar taken from the sum, this remainder 
divided by ^ would be equal to the reciprocal of | of a dollar. 






NOTATION AND NUMERATION. Itf 

DECIMAL FRACTIONS. 

NOTATION AND NUMERATION. 

202. A Decimal Fraction is one or more of the decimal 
divisions of a unit. 

Notes. — 1. The word decimal is derived from the Latin decern, which signi- 
fies ten. 

2. Decimal fractions are commonly called decimals. 

303. In the formation of decimals, a simple unit is divided 
into ten equal parts, forming decimal units of the first order, or 
tenths, each tenth is divided into ten equal parts, forming decimal 
units of the second order, or hundredths ; and so on, according to 
the following 

TABLE OF DECIMAL UNITS. 

1 single unit equals 10 tenths ; 
1 tenth " 10 hundredths; 

1 hundredth " 10 thousandths ; 
1 thousandth " 10 ten thousandths, 
etc. etc. 

204. In the notation of decimals it is not necessary to employ 
denominators as in common fractions; for, since the different 
orders of units are formed upon the decimal scale, the same law 
of local value as governs the notation of simple integral numbers, 
(57), enables us to indicate the relations of decimals by place 
or position. 

20o. The Decimal Sign (.) is always placed before decimal 
figures to distinguish them frtfm integers. It is commonly called 
the decimal point. When placed between integers and decimals 
in the same number, is sometimes called the separatrix. 

206. The law of local value, extended to decimal units, as- 
signs the first place at the right of the decimal sign to tenths ; 
the seoond, to hundredths; the third, to thousandths; and so on, 
as shown in the following 



U8 DECIMALS. 



DECIMAL NUMERATION TABLE. 

DO 

573 2 7 54.573256 

©©©©^^S © £l & ^ <*> ^ 
ooooggg gggooo. 

^pjd^^^^-M -£_ (r _,Jdrdrd 

-+3 -+3 -+3 -M T5 ~ 02 00 ~ r O -+i> -^3 -+3 

NO lO^CO (MH H W COt)< IOO 

Integers. Decimals. 

SOT. The denominator of a decimal fraction, when expressed, 
is necessarily 10, 100, 1000, or some power of 10. By examining 
the table it will be seen, that the number of places in a decimal is 
equal to the number of ciphers required to express its denomi- 
nator. Thus, tenths occupy the first place at the right of units, 
and the denominator of jL has one cipher; hundredths in the 
table extend two places from units, and the denominator of T ^ 
has two ciphers ; and so on. 

208. A decimal is usually read as expressing a certain number 
of decimal units of the lowest order contained in the decimal. 
Thus, 5 tenths and 4 hundredths, or .54, may be read, fifty-four 
hundredths. For, r % + T ^ = ffc, 

309. From the foregoing explanations and illustrations we 
derive the following 

PRINCIPLES OF DECIMAL NOTATION AND NUMERATION. 

I. Decimals are governed by the same law of local value that 
governs the notation of integers. 

II. The different orders of decimal units decrease from left to 
right, and increase from right to left, in a tenfold ratio. 



NOTATION AND NUMERATION. HQ 

III. The value of any decimal figure depends upon the place 
it occupies at the right of the decimal sign. 

IV. Prefixing a cipher to a decimal diminishes its value ten- 
fold, since it removes every decimal figure one place to the right. 

V. Annexing a cipher to a decimal does not alter its value, 
since it does not change the place of any figure in the decimal. 

VI. The denominator of a decimal, when expressed, is the 
unit, 1, with as many ciphers annexed as there are places in the 
decimal. 

VII. To read a decimal requires two numerations ) first, from 
units, to find the name of the denominator ; second, towards units, 
to find the value of the numerator. 

310. Having analyzed all the principles upon which the 
writing and reading of decimals depend, we will now present these 
principles in the form of rules. 

RULE FOR DECIMAL NOTATION. 

I. Write the decimal the same as a whole number, placing 
ciphers in the place of vacant orders, to give each significant figure 
its true local value. 
. II. Place the decimal point before the first figure. 

RULE TOR DECIMAL NUMERATION. 

I. Numerate from the decimal point, to determine the denomi- 
nator. 

II. Numerate towards the decimal point, to determine the nu- 
merator. 

III. Read the decimal as a whole number, giving it the name 
of its lowest decimal unit, or right hand figure. 

EXAMPLES FOR PRACTICE. 

Express the following decimals by figures, according to the 
decimal notation. 

1. Five tenths. Ans. .5. 

2. Thirty-six hundredths. Ans. .36. 

3. Seventy-five ten-thousandths. Ans. .0075. 



120 



DECIMALS. 



4. Four hundred ninety-six thousandths. 

5. Three hundred twenty-five ten-thousandths. 

6. One millionth. 

7. Seventy-four ten-millionths. 

8. Four hundred thirty-seven thousand five hundred forty- 
nine millionths. 

9. Three million forty thousand ten ten-millionths. 

10. Twenty-four hundred-millionths. 

11. Eight thousand six hundred forty-five hundred-thousandths. 

12. Four hundred ninety-five million seven hundred five thou- 
sand forty-eight billionths. 

13. Ninety-nine thousand nine ten-billionths. 

14. Four million seven hundred thirty-five thousand nine hun- 
dred one hundred-millionths. 

15. One trillionth. 

16. One trillion one billion one million one thousand one ten- 
trillionths. 

17. Eight hundred forty-one million five hundred sixty-three 
thousand four hundred thirty-six trillionths. 

18. Nine quintillionths. 

Express the following fractions and mixed numbers decimally : 

46.4. 



19. T %. 


Am. .3. 


25. 


46 T 4 5 . Am. 4 


20. m- 

91 i i 

99 85 

OQ 100004 
L0 - T000(5(J(J- 
94 7 04 
^- TfiOO(5<nJ5- 




26. 

27. 
28. 
29. 
30. 


fift 3 6 

30010001 ooi 

OUV T0U0"Ti(5B0- 


Read the following numbers : 






31. .24. 




38. 


8.25. 


32. .075. 




39. 


75.368. 


33. .503. 




40. 


42.0637. 


34. .00725. 




41. 


8.0074. 


35. .40000004. 




42. 


30.4075. 


36. .0000256. 




43. 


26.00005. 


37. .0010075. 




44. 


100.00000001. 



REDUCTION. 121 

REDUCTION. 
CASE I. 

211. To reduce decimals to a common denominator* 
1. Reduce .5, .24, .7836 and .875 to a common denominator. 

operation. Analysis. A common denominator must contain 

5000 as man y decimal places as is equal to the greatest 

2400 number of decimal figures in any of the given deci- 

►7g3g mals. We find that the third number contains four 

.3750 decimal places, and hence 10000 must be a common 

denominator. As annexing ciphers to decimals does 

lot alter their value, we give to each number four decimal places, by 

annexing ciphers, and thus reduce the given decimals to a common 

denominator. Hence, 

Rule. Give to each number the same number of decimal 
places, by annexing ciphers. 

Notes. — 1. If the numbers be reduced to the denominator of that one of the 
given numbers having the greatest number of decimal places, they will have 
their least common decimal denominator. 

2. An integer may readily be reduced to decimals by placing the decimal 
point after units, and annexing ciphers ; one cipher reducing it to tenths, two 
ciphers to hundredths, three ciphers to thousandths, and so on. 

EXAMPLES FOR PRACTICE. 

1. Reduce .18, .456, .0075, .000001, .05, .3789, .5943786, and 
.001 to their least common denominator. 

2. Reduce 12 thousandths, 185 millionths, 936 billionths, and 
7 trillionths to their least common denominator. 

3. Reduce 57.3, 900, 4.7555, and 100.000001 to their least 
common denominator. 

CASE II. 

212. To reduce a decimal to a common fraction. 
1. Reduce .375 to an equivalent common fraction. 

. ". _„ Analysis. Writing the decimal 

OPERATION. ° 

375 3 figures, .375, over the common de- 

• 61b = jo oo = «• nominator, 1000, we have T %\% = f . 
Hence, 

11 



122 DECIMALS. 

Rule. Omit the decimal point, and supply the proper denomi- 
nator. 

EXAMPLES FOR PRACTICE. 

1. Reduce .75 to a common fraction. Ans. |. 

2. Reduce .625 to a common fraction. Ans. f . 

3. Reduce .12 to a common fraction. 

4. Reduce .68 to a common fraction. 

5. Reduce .5625 to a common fraction. 

6. Reduce .024 to a common fraction. Ans. T |g. 

7. Reduce .00032 to a common fraction. Ans. sy 1 ^. 

8. Reduce .002624 to a common fraction. Ans. j^\s- 

9. Reduce 13| to a common fraction. 

OPERATION. 

IS 1 

191 ZZJ 4 2 

Note. — The decimal .13J may properly be called a complex decimal. 

10. Reduce .57^ to a common fraction. Ans. |. 

11. Reduce .66| to a common fraction. Ans. f . 

12. Reduce .444| to a common fraction. 

13. Reduce .024| to a common fraction. Ans. jfj^- 

14. Reduce .984| to a common fraction. 

15. Express 7.4 by an integer and a common fraction. 

Ans. 7|. 

16. Express 24.74 by an integer and a common fraction. 

17. Reduce 2.1875 to an improper fraction. Ans. f §. 

18. Reduce 1.64 to an improper fraction. 

19. Reduce 7.496 to an improper fraction. Ans. f ||. 

CASE III. 

213. To reduce a common fraction to a decimal. 
1. Reduce f to its equivalent decimal. 

first operation. Analysis. We first annex 

5 5000- 625 aqp; * ne same number of ciphers to 

•° ° both terms of the fraction; this 



5 ~ S0U0 — Ttf <5ZJ 



does not alter its value, (174, 



REDUCTION. 123 

second operation. Ill) ; we then divide both re- 

8 ) 5.000 suiting terms by 8, the signifi- 

cant figure of the denominator, 

625 

and obtain the decimal denom- 
inator, 1000. Omitting the denominator, and prefixing the sign, we 
have the equivalent decimal, .625. 

In the second operation, we omit the intermediate steps, and obtain 
the result, practically, by annexing the three ciphers to the nume- 
rator, 5, and dividing the result by the denominator, 8. 

2. Reduce T |^ to a decimal. 

operation. Analysis. Dividing as in the former ex- 

125 ) 3.000 ample, we obtain a quotient of 2 figures, 24. 

024 Rut since 3 ciphers have been annexed to the 

numerator, 3, there must be three places in the 

required decimal ; hence we prefix 1 cipher to the quotient figures, 

24. The reason of this is shown also in the following operation. 

125 125000 — 1000 — • V * J ^ t 

214. From these illustrations we derive the following 
Rule. I. Annex ciphers to the numerator, and divide by the 

denominator. 

II. Point off as many decimal places in the result as are equal 

to the number of ciphers annexed. 

Note. — If the division is not exact when a sufficient number of decimal 
figures have been obtained, the sign, +, may be annexed to the decimal to indi- 
cate that there is still a remainder. When this remainder is such that the next 
decimal figure would be 5 or greater than 5, the last figure of the terminated 
decimal may be increased by 1, and the sign, — , annexed. And in general, + 
denotes that the written decimal is too small, and — denotes that the written 
decimal is too large ,• the error always being less than one half of a unit in the 
last place of the decimal. 

EXAMPLES FOR PRACTICE. 

1. Reduce f to a decimal. Ans. .75. 

2. Reduce T 5 g to a decimal." Ans. .315. 

3. Reduce g to a decimal. 

4. Reduce if to a decimal. 

5. Reduce {§ to a decimal. 

6. Reduce ^ to a decimal. Ans. .04. 

7. Reduce ^^ to a decimal. Ans. .068. 



124 DECIMALS. 

8. Reduce ^f to a decimal. Ans. .59375 

9. Reduce y^oo to a ^ ec i ma l- i 

10. Reduce ^ to a decimal. Ans. .29167 — . 

11. Reduce y 9 ^ to a decimal. 

12. Reduce f § to a decimal. Ans. .767857+. 

13. Reduce 7| to the decimal form. Ans. 7.125. 

14. Reduce 56^ to the decimal form. Ans. 66.078125. 

15. Reduce 32| to the decimal form. 

16. Reduce .24^ to a simple decimal. 

17. Reduce 5.78|| to a simple decimal. 

18. Reduce .3 t Hq to a simple decimal. Ans. .30088. 

19. Reduce 4.0 2 2 ^ to a simple decimal. Ans. 4.008. 

20. Reduce -30^$^ to a simple decimal. 



ADDITION. 

S215. Since the same law of local value extends both to the 
right and left of units' place ; that is, since decimals and simple 
integers increase and decrease uniformly by the scale of ten, it is 
evident that decimals may be added, subtracted, multiplied and 
divided in the same manner as integers. 

316. 1. What is the sum of 4.75, .246, 37.56 and 12.248 ? 

operation. Analysis. We write the numbers so that units of 

4.75 like orders, whether integral or decimal, shall stand 

.246 in the same columns ; that is, units under units, tenths 

37.56 under tenths, etc. This brings the decimal points 

12.248 directly under each other. Commencing at the right 

54 804 hand, we add each column separately, carrying 1 for 

every ten, according to the decimal scale ; and in the 
result we place the decimal point between units and tenths, or directly 
under the decimal points in the numbers added. Hence the fol- 
lowing 

Rule. I. Write the numbers so that the decimal points shall 
stand directly under each other. 

II. Add as in whole numbers, and place the decimal point, in 
the result, directly under the points in the numbers added. 



ADDITION. 125 

EXAMPLES FOR PRACTICE. 

1. Add .375, .24, .536, .78567, .4637, and .57439. 

Ans. 2.97476. 

2. Add 5.3756, 85.473, 9.2, 46.37859, and 45.248377. 

Ans. 191.675567. 

3. Add .5, .37, .489, .6372, .47856, and .02524. 

4. Add .46f , .3251, .16^, and .275 T V Ans. 1.2296625. 

5. Add 4.6£, 7.32 3 \, 5.3784*, and 2.64878f . 

6. Add 4.3785, 2|, 5f , and 12.4872. Ans. 24.19609+. 

7. What is the sum of 137 thousandths, 435 thousandths, 836 
thousandths, 937 thousandths, and 496 thousandths ? 

Ans. 2.841. 

8. What is the sum of one hundred two ten-thousandths, thir- 
teen thousand four hundred twenty-six hundred thousandths, five 
hundred sixty-seven millionths, three millionths, and twenty-four 
thousand seven hundred-thousandths ? 

9. A farm has five corners; from the first to the second is 34.72 
rods; from the second to the third, 48.44 rods; from the third to 
the fourth, 152.17 rods; from the fourth to the fifth, 95.36 rods; 
and from the fifth to the first, 56.18 rods. What is the whole 
distance around the farm ? 

10. Find the sum of §4, ^% f j^q, and j^-%% in decimals, correct 
to the fourth place. Ans. 6658 + . 

Note. — In the reduction of each fraction, carry the decimal to at ,^sst the 
fifth place, in order to insure accuracy in the fourth place. 

11. A man owns 4 city lots, containing 16 T 6 ^ rods, 15^ rods, 
18|f rods, and 14^ rods of land, respectively; how many rods 
in all? 

12. What is the sum of 4^ decimal units of the first order, 2| 
of the second order, 9| of the third order, and 3^ of the fourth 
order? Ans. .486929. 

13. What is the approximate sum of 1 decimal unit of the first 
order, ^ of a unit of the second order, ^ of a unit of the third 
order, | of a unit of the fourth order, i of a unit of the fifth order, 
£ of a unit of the sixth order, and ^ of a unit of the seventh order ? 

Ans. .1053605143—. 
11* 



126 



DECIMALS. 



SUBTRACTION. 
S17. I- From 4.15G take .5783. 

OPERATION. 

4. 1 560 Analysis. Wo write the given numbers as in addi- 

■57cv> tion, reduce the decimals to a common denominator, 

3.5777 5,n( l Hubfcract as in integers. Or, we may, in practice, 

( omit the ciphers necessary to reduce the decimals to a 

common denominator, and merely conceive them to be 

4.150 annexed, subtracting as otherwise. Hence the fol- 

.5783 lowing 

1.5777 

Rule. I. Write the numbers so that the decimal points shall 
stand directly under each other, 

II. Subtract as in whole numbers, and place the decimal point 
in the result directly under the points in the given numbers. 

EXAMPLES FOR PRACTICE. 

(1.) (2.) (3.) 

Mimiond, .0876 48.3676 36.5 

subtrahend, .3598 23.98 35.875632 

Remainder, ^6278 24.3876 .624368 

4. From 37.456 take 24.367. Ans. 13.089. 

5. From 1.0066 take .15. 

6. From 1000 take .001. Ans. 999.999. 

7. From 36| take 22£§. Ans. 14.27. 

8. From .56g take .55[§f. 

9. From 7 J take 5,V Ans. 1.7708+. 

10. From ;;;;> takeoff 

11. From one take one trillionth. Ans. .999099999999. 

12. A speculator baying 57436 acres of land, sold at different 
times 586.74 aores, 1756.19 acres, 3678.47 acres, 0572.15 acres, 
7586.59 acres, and 4785.94 acres; how much land has he 
remaining? 

18. Kind the difference between f$$f£ and lff|f, correct to 
the fifth decimal place. Ans. 4.17298+ . 



M ULTIPLICATION. J27 



MULTIPLICATION. 



218. In multiplication of decimals, the location of the decimal 
point in the product depends upon the following principles : 

I. The number of ciphers in the denominator of a decimal is 
equal to the number of decimal places, (200, VI). 

II. If two decimals, in the fractional form, be multiplied to- 
gether, the denominator of the product must contain as many 
ciphers as there are decimal places in both factors. Therefore, 

III. The product of two decimals, expressed in the decimal 
form, must contain as many decimal places as there are decimals 
in both factors. 

1. Multiply .45 by .7. 

operation. Analysis. We first multiply 

45 as in whole numbers ; then, 

7 since the multiplicand has 2 

~ttt decimal places and the multi- 

plier 1, we point off 2 + 1 = 3 
ritooF. decimal places in the product, 

lV(J X j\ = tVtAj = .315 (HI). The reason of this is 

further illustrated in the proof, 
a method applicable to all similar cases. 

310. Hence the following 

Rule. Multiply as in whole numbers, and from the right hand 

of the product point off as many figures for decimals as there are 

decimal places in both factors. 

Notes. — 1. If there be not as many figures in the product as there are deci- 
mals in both factors, supply the deficiency by prefixing ciphers. 

2. To multiply a decimal by 10, 100, 1000, etc., remove the point as many 
places to the right as there are ciphers on the right of the multiplier. 

EXAMPLES FOR PRACTICE. 

1. Multiply .75 by .41. Ans. .3075. 

2. Multiply .436 by .24. 

3. Multiply 5.75 by .35. Ans. 2.0125. 

4. Multiply .756 by .025. Ans. .019. 

5. Multiply 3.784 by 2.475. 



128 DECIMALS. 

6. Multiply 7.23 by .0156. Ans. .112788. 

7. Multiply .0075 by .005. Ans. .0000375. 

8. Multiply 324 by .324. 

9. Multiply 75.64 by .225. 

10. Multiply 5.728 by 100. Ans. 572.8. 

11. Multiply .36 by 1000. 

12. Multiply .000001 by 1000000. 

13. Multiply .576 by 100000. 

14. Multiply 7f by 5*. Ans. 42.625. 

15. Multiply .63| by 24. 

16. Multiply 4 T 5 g by 7^. Ans. 31.74. 

17. Find the value of 3.425 x 1.265 x 64. Ans. 277.288. 

18. Find the value of 32 x .57825 x .25. 

19. Find the value of 18.375 x 5.7 x 1.001. 

Ans. 104.8422375. 

20. If a cubic foot of granite weigh 168.48 pounds, what will 
be the weight of a granite block 4 feet long, 3 feet wide, and 2 
feet thick ? 

21. When a bushel of corn is worth 2.8 bushels of oats, how 
many bushels of oats must be given in exchange for 36 bushels 
of corn and 48 bushels of oats ? Ans. 148.9. 



CONTRACTED MULTIPLICATION. 

220* To obtain a given number of decimal places in 
the product. 

It is frequently the case in multiplication, that a greater number 
of decimal figures is obtained in the product, than is necessary 
for practical accuracy. This may be avoided by contracting each 
partial product to the required number of decimal places. 

To investigate the principles of this method, let us take the two 
decimals .12345 and .54321, and having reversed the order of the 
digits in the latter, and written it under the former, multiply each 
figure of the direct number by the figure below in the reversed num- 
ber, placing the products with like orders of units in the same column, 
thus : 



CONTRACTED MULTIPLICATION. J29 

.12345 direct = .12345 
.54321 reversed = 12345. 



.000025 = .00005 X .5 
.000016 = .0004 X .04 
.000009 == .003 X .003 
.000004 = .02 X .0002 
.000001 = .1 X .00001. 

In this operation we perceive that all the products are of the same 
order ; and this must always be, whether the numbers used be frac- 
tional, integral, or mixed. For, as we proceed from right to left in 
the multiplication, we pass regularly from lower to higher orders in 
the direct number, and from higher to lower in the reversed number. 
Hence 

221. If one number be written under another with the order 
of its digits reversed, and each figure of the reversed number be 
multiplied by the figure above it in the direct number, the prod- 
ucts will all be of the same order of units. 

1. Multiply 4.78567 by 3.25765, retaining only 3 decimal 
places in the product. 

operation. Analysis. Since the product 

J. 7ft ^£7 °^ an y % ure by units is of the 

WlW 3 same order as the figure multi- 

plied, (82, II,) we write 3, the 
units of the multiplier, under 
5, the third decimal figure of 
the multiplicand, and the lowest 
order to be retained in the pro- 
duct ; and the other figures of 
15.589 =fc, Ans. the mu i t i p ii er we wr i te j n tne 

inverted order, extending to the left. Then, since the product of 3 
and 5 is of the third order, or thousandths, the products of the other 
corresponding figures at the left, 2 and 8, 5 and 7, 7 and 4, etc., will 
be thousandths ; and we therefore multiply each figure of the 
multiplier by the figures above and to the left of it in the multipli- 
cand, carrying from the rejected figures of the multiplicand, as fol- 
lows : 3 times 6 are 18, and as this is nearer 2 units than one of the 
next higher order, we must carry 2 to the first contracted product ; 3 
times 5 are 15, and 2 to be carried are 17 ; writing the 7 under the 
3, and multiplying the other figures at the left in the usual manner, 

I 



14357 = 


4785 


X 


3 + 2 


957 = 


478 


X 


2 + 1 


239 = 


47 


X 


5 + 4 


33 = 


4 


X 


7 + 5 


3 = 





X 


6 + 3 



130 DECIMALS. 

we obtain 14357 for the first partial product. Then, beginning with 
the next figure of the multiplier, 2 times 5 are 10, which gives 1 to 
be carried to the second partial product ; 2 times 8 are 16, and 1 to be 
carried are 17 ; writing the 7 under the first figure of the former pro- 
duct, and multiplying the remaining left-hand figures of the mul- 
tiplicand, we obtain 957 for the second partial product. Then, 5 
times 8 are 40, which gives 4 to be carried to the third partial pro- 
duct ; 5 times 7 are 35 and 4 are 39 ; writing the 9 in the first column 
of the products, and proceeding as in the former steps, we obtain 239 
for the third partial product. Next, multiplying by 7 in the same 
manner, we obtain 33 for the fourth partial product. Lastly, begin- 
ning 2 places to the right in the multiplicand, 6 times 7 are 42 ; 6 
times 4 are 24, and 4 arc 28, which gives 3 to be carried to the fifth 
partial product; 6 times is 0, and 3 to be carried are 3, wliich we 
write for the last partial product. Adding the several partial pro- 
ducts, and pointing off 3 decimal places, we have 15.589, the required 
product. 

222. From these principles and illustrations we derive the 
following 

Rule. I. Write the multiplier with the order of its figures 
reversed , and with the units 7 place under that figure of the multi- 
plicand which is the lowest decimal to be retained in the product. 

II. Find the product of each figure of the multiplier by the 
figures above and to the left of it in the multiplicand y increasing 
each partial product by as many units as would have been carried 
from the rejected part of the multiplicand, and one more when the 
highest figure in the rejected part of any product is 5 or greater 
than 5 ; arid write these partial pQ*oducts with the lowest figure of 
each in the same column. 

III. Add the partial products, and from the right hand of the 

result point off the required number of decimal figures. 

Notes. — 1. In obtaining the number to be carried to each contracted partial 
product, it is generally necessary to multiply (mentally) only one figure at the 
right of the figure above the multiplying figure; but when the figures are large, 
the multiplication should commence at least two places to the right. 

2. Observe, that when the number of units in the highest order of the rejected 
part of the product is between 5 and 15, carry 1 ; if between 15 and 25 carry 
2; if between 25 and 35 carry 3 ; and so on. 

3. There is always a liability to an error of one or two units in the last place; 
and as the answer may be either too great or too small by the amount of this 



CONTRACTED MULTIPLICATION. JgJ 

error, the uncertainty may be indicated by the double sign, ±, read, plus, or 
minus, and placed after the product. 

4. When the number of decimal places in the multiplicand is less than the 
number to be retained in the product, supply the deficiency by annexing ciphers. 

EXAMPLES FOR PRACTICE. 

1. Multiply 236.45 by 32.46357, retaining 2 decimal places, 
and 2.563789 by .0347263, retaining 6 decimal places in the 
product. 

OPERATION. OPERATION. 

236.450 2.563789 

75364.23 362 7430. 

709350 76914 

47290 10255 

9458 1795 

1419 51 

71 15 

12 1 



2 .089031 ± 



7676.02 =fc 

2. Multiply 36.275 by 4.3678, retaining 1 decimal place in the 
product. Ans. 158,4 dz. 

3. Multiply .24367 by 36.75, retaining 2 decimal places in the 
product. 

4. Multiply 4256.785 by .00564, rejecting all beyond the third 
decimal place in the product. Ans. 24.008 =fc. 

5. Multiply 357.84327 by 1.007806, retaining 4 decimal places 
in the product. 

6. Multiply 400.756 by 1.367583, retaining 2 decimal places in 
the product. Ans. 548.07 =b. 

7. Multiply 432.5672 by 1.0666666, retaining 3 decimal places 
in the product. 

8. Multiply 48.4367 by 2^, extending the product to three 
decimal places. Ans. 103.418 =b. 

9. Multiply 7 T f 3 by 3|||, extending the product to three 
decimal places. 

10. The first satellite of Uranus moves in its orbit 142.8373 + 



Jg2 DECIMALS. 

degrees in 1 day; find how many degrees it will move in 2.52035 
days, carrying the answer to two decimal places. 

Ans. 360.00 degrees. 

11. A gallon of distilled water weighs 8.33888 pounds ; how 
many pounds in 35.8756 gallons? Ans. 299.16 db pounds. 

12. One French metre is equal to 1.09356959 English yards; 
how many yards in 478.7862 metres. Ans. 523.58 ± yards. 

13. The polar radius of the earth is 6377397.6 metres, and the 
equatorial radius, 6356078.96 metres ; find the two radii, and their 
difference, to the nearest hundredth of a mile, 1 metre being equal 
to 0.000621346 of a mile. 



DIVISION. 

223. In division of decimals the location of the decimal 
point in the quotient depends upon the following principles : 

I. If one decimal number in the fractional form be divided by 
another also in the fractional form, the denominator of the quotient 
must contain as many ciphers as the number of ciphers in the de- 
nominator of the dividend exceeds the number in the denominator 
of the divisor. Therefore, 

II. The quotient of one number divided by another in the deci- 
mal form must contain as many decimal places as the number of 
decimal places in the dividend exceed the number in the divisor. 

1. Divide 34.368 by 5.37. 

operation. Analysis. We first divide as 

p\ 37 ^ 34 3fi£ f ft 1 * n wno ^ e numbers ; then, since the 

32 22 dividend has 3 decimal places and 

— the divisor 2, we point off 3 — 2 

^ *^" = 1 decimal place in the quotient, 

(II). The correctness of the work 
is shown in the proof, where the 
dividend and divisor are written as 
n M common fractions. For, when we 

3_43_6 8 y 100 64 ft A _ ' 

loou ~ ^37 ill v# ~ have canceled the denominator of 

the divisor from the denominator 
of the dividend, the denominator of the quotient must contain as 



PROOF. 



DIVISION. X33 

many ciphers as the number in the dividend exceed those in the 
divisor. 

224L Hence the following 

Rule. Divide as in whole numbers, and from the right hand 

of the quotient point off as many places for decimals as the decimal 

places in the dividend exceed those in the divisor. 

Notes.— 1. If the number of figures in the quotient be less than the excess of 
the decimal places in the dividend over those in the divisor, the deficiency must 
be supplied by prefixing ciphers. 

2. If there be a remainder after dividing the dividend, annex ciphers, and 
continue the division : the ciphers annexed are decimals of the dividend. 

3. The dividend should always contain at least as many decimal places as the 
divisor, before commencing the division ,• the quotient figures will then be inte- 
gers till all the decimals of the dividend have been used in the partial dividends. 

4. To divide a decimal by 10, 100, 1000, etc., remove the point as many places 
to the left as there are ciphers on the right of the divisor. 

EXAMPLES FOR PRACTICE. 

1. Diyide 9.6188 by 3.46. Ans. 2.78. 

2. Divide 46.1975 by 54.35. Ans. .85. 

3. Divide .014274 by .061. Ans. .234. 

4. Divide .952 by 4.76. 

5. Divide 345.15 by .075. Ans. 4602. 

6. Divide .8 by 476.3. Ans. .001679+. 

7. Divide .0026 by .003. 

8. Divide 3.6 by .00006. Ans. 6000. 

9. Divide 3 by 450. 

10. Divide 75 by 10000. 

11. Divide 4.36 by 100000. 

12. Divide .1 by .12, 

13. Divide 645.5 by 1000. 

14. If 25 men build 154.125 rods of fence in a day, how much 
does each man build ? 

15. How many coats can be made from 15.34 yards of cloth, 
allowing 2.7 yards for each coat? 

16. If a man travel 36.34 miles a day, how long will it take 
him to travel 674 miles ? Ans. 18.547+days. 

17. How many revolutions will a wheel 14.25 feet in circum- 
ference make in going a distance of 1 mile or 5280 feet ? 

12 



134 



DECIMALS. 



CONTRACTED DIVISION. 

225. To obtain a given number of decimal places in 
the quotient. 

In division, the products of the divisor by the several quotient 
figures maybe contracted, as in multiplication, by rejecting at each 
step the unnecessary figures of the divisor, (220). 

1. Divide 790.755197 by 32.4687, extending the quotient to 
two decimal places. 

FIRST CONTRACTED METHOD. COMMON METHOD. 

32.4687 ) 790.755197 ( 24.35 
649 4 



32.4687)790.7 55198(24.35 



649 3 



74 



1413 




129 9 




114 




97 




17 




16 




1 




SECOND CONTRACTED METHOD 


32.4687 ) 790.755197 
53.42 141 3 




114 




17 



1413 
129 8 



811 

748 



U5 
97 



17 

16 



0639 
4061 



65787 
23435 



1 42352 
Analysis. In the first method 
of contraction, we first compare the 
3 tens of the divisor with the 79 
tens of the dividend, and ascertain 
that there will be 2 integral places 
in the quotient ; and as 2 decimal 
L places are required, the quotient 

must contain 4. places in all. Then 
assuming the four left hand figures of the divisor, we say 3246 is con- 
tained in 7907, 2 times ; multiplying the assumed part of the divisor 
by 2, and carrying 2 units from the rejected part, as in Contracted 
Multiplication of Decimals, we have 6494 for the product, which sub- 
tracted from the dividend, leaves 1413 for a new dividend. Now, 
since the next quotient figure will be of an order next below the 
former, we reject one more place in the divisor, and divide by 324, 
obtaining 4 for a quotient, 1299 for a product, and 114 for a new divi- 
dend. Continuing this process till all the figures of the divisor are 



CONTRACTED DIVISION. jgg 

rejected, we have, after pointing. off 2 decimals as required, 24.35 for 
a quotient. Comparing the contracted with the common method, we 
see the extent of the abbreviation, and the agreement of the corres- 
ponding intermediate results. 

In the second method of contraction, the quotient is written with 
its first figure under the lowest order of the assumed divisor, and the 
other figures at the left in the reverse order. By this arrangement, 
the several products are conveniently formed, by multiplying each 
quotient figure by the figures above and to the left of it in the divisor, 
by the rule for contracted multiplication, (222), and the remainders 
only are written as in (112). 

226. From these illustrations we derive the following 
Rule. I. Compare the highest or left hand figure of the divisor 
with the units of like order in the dividend, and determine how 
many figures will he required in the quotient. 

II. For the first contracted divisor, take as many significant 
figures from the left of the given divisor as there are places re- 
quired in the quotient ; and at each subsequent division reject one 
'place from the right of the last preceding divisor. 

III. In multiplying by tJie several quotient figures, carry from 
the rejected figures of the divisor as in contracted multiplication. 

Notes. — 1. Supply ciphers, at the right of either divisor or dividend, when 
necessary, before commencing the work. 

2. If the first figure of the quotient is written under toe lowest assumed figure 
of the divisor, and the other figures at the left in the inverted order, the several 
products will be formed with the greatest convenience, by simply multiplying 
each quotient figure by the figures above and to the left of it in the divisor. 



EXAMPLES FOR PRACTICE. 

1. Divide 27.3782 by 4.3267, extending the quotient to 3 deci- 
mal places. Arts. 6.328 db. 

2. Divide 487.24 by 1.003675, extending the quotient to 2 
decimal places. 

3. Divide 8.47326 by 75.43, extending the quotient to 5 deci- 
mal places. 

4. Divide .8487564 by .075637, extending the quotient to 3 
decimal places. Ans. 11.221 ±. 



136 DECIMALS. 

5. Divide 478.325 by 1-43|, extending the quotient to 3 deci- 
mal places. Ans. 332.942 zk. 

6. Divide 8972.436 by 756.3452, extending the quotient to 4 
decimal places. 

7. Divide 1 by 1.007633, extending the quotient to 6 decimal 
places. Ans. .992425 ±. 

8. Find the quotient of .95372843 divided by 44.736546, true 
to 8 decimal places. 

9. Keduce f § f f to a decimal of 4 places. Ans. .7448 ±. 

CIRCULATING DECIMALS. 

227. Common fractions can not always be exactly expressed in 
the decimal form; for in some instances the division will not be 
exact if continued indefinitely. 

228. A Finite Decimal is a decimal which extends a limited 
number of places from the decimal point. 

229. An Infinite Decimal is a decimal which extends an 
unlimited number of places from the decimal point. 

230. A Circulating Decimal is an infinite decimal in which 
a figure or set of figures is continually repeated in the same order • 
as .3333+, or .437437437+. 

231. A Repetend is the figure or set of figures continually 
repeated When a repetend consists of a single figure, it is in- 
dicated by a point placed over it; when it consists of more than 
one figure, a point is placed over the first, and one over the last 
figure. Thus, the circulating decimals .55555+ and .324324324+ 
are written, .3 and .324. 

232. A repetend is said to be expanded when its figures are 
continued in their proper order any number of places toward the 
right; thus, .24, expanded is .2424+, or .242424242+. 

233. Similar Eepetends are those which begin at the same 
decimal place or order; as .37, and .5, .24 and .375, 1.56 and 24.3. 

234. Conterminous Repetends are those which end at the 
same decimal place or order; as .75 and 1.53, .567, and 3.245. 

Note —Two or more repetends are Similar and Conterminous when they begin 
and end at the same decimal places or orders. 



CIRCULATING DECIMALS. 137 

2*1*5. A Pure Circulating Decimal is one which contains no 
figures but the repetend; as .7, or .704. 

236. A Mixed Circulating Decimal is one which contains 
other figures ; called finite places, before the repetend j as .54, or 
.013245, in which .5 and .01 are the finite places. 

PROPERTIES OF FINITE AND CIRCULATING DECIMALS. 

237. The operations in circulating decimals depend upon the 

following properties. 

Note. — 1. The common fractions referred to are understood to be proper frac- 
tions, in their lowest terms, 

I. Every fraction whose denominator contains no other prime 
factor than 2 or 5 will give rise to a finite decimal ; and the num- 
ber of decimal places will be equal to the greatest number of equal 
factors, 2 or 5, in the denominator. 

For, in the reduction, every cipher annexed to the numerator mul- 
tiplies it by 10, or introduces the two prime factors, 2 and 5, and also 
gives 1 decimal place in the result. Hence the division will be exact 
when the number of ciphers annexed, or the number of decimal 
places obtained, shall be equal to the greatest number of equal factors, 
2 or 5, to be canceled from the denominater. 

II. Every fraction whose denominator contains any other prime 
factor than 2 or 5, will give rise to an infinite decimal. 

For, annexing ciphers to the numerator introduces no other prime 
factors than 2 and 5 ; hence the denominator will always be prime to 
the numerator, and the division can never be exact. 

III. Every infinite decimal derived from a common fraction is 
also a circulating decimal; and the number of places in the 
repetend must be less than the number of units in the denominator 
of the common fraction. 

For, in every division, the number of possible remainders is limited 
to the number of units in the divisor, less 1 ; thus, in dividing by 7, 
the only possible remainders are 1, 2, 3, 4, 5, and 6. Hence, in the 
reduction of a common fraction to a decimal, some of the remainders 
must repeat before the number of decimal places obtained equals the 
number of units in the denominator ; and this will cause the inter- 
mediate quotient figures to repeat. 
12* 



138 DECIMALS. 

Notes. — 2. It will be found that the number of places in the repetend is always 
equal to the denominator less 1, or to some factor of this number. Thus, the 
repetend arising from f has 7 — 1 = 6 places ; the repetend arising from f f has 
41-1 = 5 places. 

3. A 'perfect repetend is one which consists of as many places, less 1, as there 
are units in the denominator of the equivalent fraction. 

4. If the denominator of a fraction contains neither of the factors 2 and 5, it 
will give rise to a pure repetend. But if a circulating decimal is derived from a 
fraction whose denominator contains either of the factors 2 or 5, it will contain 
as many finite places as the greatest number of equal factors 2 or 5 in the de- 
nominator. 

IV. If to any number we annex as many ciphers as there are 
places in the number, or more, and divide the result by as many 
9's as the number of ciphers annexed, both the quotient and re- 
mainder will be the same as the given number. 

For, if we take any number of two places, as 74, and annex two 
ciphers, the result divided by 100 will be equal to 74 ; thus, 
7400 -7- 100 = 74. 
Now subtracting 1 from the divisor, 100, will add as many units 
to the quotient, 74, as the new divisor, 99, is contained times in 74, 
(115, II) ; thus, 

7400 + 99 = 74 + ft, or 74ft; 

that is, if two ciphers be annexed to 74, and the result be divided by 
99, both quotient and remainder will be 74. In like manner, annex- 
ing three ciphers to 74, and dividing by 999, we have 

74000^-999 = 74^; 
and the same is true of any number whatever. 

V. Every pure circulating decimal is equal to a common frac- 
tion whose numerator is the repeating figure or figures, and whose 
denominator is as many 9's as there are places in the repetend. 

For, if we take any fraction whose denominator is expressed by 
some number of 9's, as f £, then according to the last property, annex- 
ing two ciphers to the numerator, and reducing to a decimal, we have 
f £ = 24f|. In like manner, carrying the decimal two places farther, 
.24f| = .2424|4 ; hence, f£ = 24. By the same principle, we have 
| =.2 ; ¥ V = .01 ; A = -02 ; ?h = .001 ; «* = •&* 5 and so on. And 
it is evident that all possible repetends can thus be derived from frac- 
tions whose numerators are the repeating figures, and whose denomi- 
nators are as many 9's as there are repeating figures. 



CIRCULATING DECIMALS. ^39 

Note 5. — It follows from the last property, that any fraction from which a pure 
repetend can be derived is reducible to a form in which the denominator is some 
number of 9's ; thus T % = lif fo| J 37 = iM- This is true of every fraction 
whose denominator terminates with 1, 3, 7, or 9. 

VI. Any repetend may be reduced to another equivalent repe- 
tend, by expanding it, and moving either the second point, or 
both points, to the right; provided that in the result they be so 
placed as to include the same number of places as are contained 
in the given repetend, or some multiple of this number. 

For, in every such reduction, the new repetend and the given repe- 
tend, when expanded indefinitely, will give results which are identical. 
Thus, .536 = .536536, or .536536536, or .5365, or .53653, or .5365365, 
or .53653653653 ; because each of these new repetends, when ex- 
panded, gives .53653653653653653653+. 

Note 6. — If in any reduction, the new repetend should not contain the same 
number of places, or some multiple of the same number, as the given repetend, 
we should not have, in the expansions, the same figures repeated in the same 
order, 

REDUCTION. 
CASE I. 

238. To reduce a pure circulating decimal to a 
common fraction. 

I. Reduce .675 to a common fraction. 

operation. Analysis. Since the repetend has 3 

675 = £ 7 ^ = ^ 5 places, we take for the denominator of 

the required fraction the number ex- 
pressed by three 9's, (237, III). Hence, 

Rule. Omit the points and the decimal sign, and write the 
figures of the repetend for the numerator of a common fraction, 
and as many 9's as there are places in the repetend for the de- 
nominator. 

examples for practice. 

1. Reduce .45 to a common fraction. Ans. T 5 T . 

2. Reduce .66 to a common fraction. 

3. Reduce .279 to a common fraction. Ans. ^j\. 



J40 DECIMALS. 

4. Reduce .423 to a common fraction. Ans. T 4 T 7 T . 

5. Reduce .923078 to a common fraction. Ans. {§. 

6. Reduce .94121 to a common fraction. 

7. Reduce 4.72 to a mixed number. Ans. 4 T 8 T . 

8. Reduce 2.279 to an improper fraction. Ans. f f . 

9. Reduce 2.97 to an improper fraction. Ans. W°. 
Note. — According to 237, VI, 2.97 = 2.972. 

10. Reduce 15.0 to a mixed number. Ans. 15^1 3. 

CASE 11. 

239. To reduce a mixed circulating decimal to a 
common fraction. 

1. Reduce .0756 to a common fraction. 

operation. Analysis. Since .756 is equal 

.0756 = jffo = -J/5 to Ul .0756 will be T V of ft$, 

nr 7 5 6 14 

or ^9 Otf — TST* 

2. Reduce .647 to a common fraction. 

operation. Analysis. Reducing the finite 

647 =- - 6 - 4 - -I- 2 P ar * an( ^ the repetend separately 

g40 g4° 7 to fractions, we have T %\ + ^fo. 

== — onn ("orwi ^° re( ^ uce these fractions to a 

yuu common denominator, we must 

_ 640 — 64+7 multiply the terms of the first by 

900 9 ; but the numerator, 64, may 

647 — 64 be multiplied by 9 by annexing 

= 900 1 cip ner an d subtracting 64 from 

583 . the result, giving — — — — , for 

= 90Q, Ans. * 5 900 ' 

the first fraction reduced. The 

^ r > numerator of the sum of the two 

.647 given decimal. fractions will therefore be 640 

64 finite figures. — 64 + 7 = 583, and supplying 

^o the common denominator, we have 

_~~ |§g. In the second operation, 

y Ans. the intermediate steps are omitted. 

900 Hence the following 

Rule. I. From the given circulating decimal subtract the finite 
party and the remainder will be the required numerator. 



CIRCULATING DECIMALS. 141 

J 
II. Write as many 9's as there are figures in the repetend, with 
as many ciphers annexed as there are finite decimal figures, for 
the required denominator. 

EXAMPLES TOR PRACTICE. 

1. Reduce .57 to a common fraction. Ans. f f . 

2. Reduce .048 to a common fraction. Ans. ^y^. 

3. Reduce .6472 to a common fraction. 

4. Reduce .6590 to a common fraction. Ans. ||. 

5. Reduce .04648 to a common fraction. Ans. -g 4 ^. 

6. Reduce .1004 to a common fraction. 

7- Reduce .9285714 to a common fraction. Ans. ||. 

8. Reduce 5.27 to a common fraction. Ans. ||. 

9. Reduce 7.0126 to a mixed number. Ans. 7^$. 

10. Reduce 1.58231707 to an improper fraction. Ans. |^|. 

11. Reduce 2.029268 to an improper fraction. 

CASE III. 

340. To make two or more repetends similar and 

conterminous. 

1. Make .47, .53675, and .37234 similar and conterminous. 

operation. Analysis. The first of 

the given repetends begins 
.47 = .47474747474747 ^ at £ e plac / of tenths ; the 

.53675 = .53675675675675 I Ans. second at the place of thou- 
.37234 = .37234723472347 J sandths, and the third at 

the place of hundredths; 
and as the points in any repetend cannot be moved to the left over 
the finite places, we can make the given repetends similar, only by 
moving the points of at least two of them to the right. 

Again, the first repetend has 2 places, the second 3 places, and the 
third 4 places ; and the number of places in the new repetends must 
be at least 12, which is the least common multiple of 2, 3, and 4. 
We therefore expand the given repetends, and place the first point in 
each new repetend over the third place in the decimal, and the second 
point over the fourteenth, and thus render them similar and conter- 
minous. Hence the following 



142 DECIMALS. 

Rule. I. Expand the repetends, and place the first point in 
each over the same order in the decimal. 

II. Place the second point so that each new repetend shall con- 
tain as many places as there are units in the least common mul- 
tiple of the number of places in the several given repetends. 

Note. — Since none of the points can be carried to the left, some of them must 
be carried to the right, so that each repetend shall have at least as many finite 
places as the greatest number in any of the given repetends. 

EXAMPLES FOR PRACTICE. 

1. Make .43, .57, .4567, and .5037 similar and conterminous. 

2. Make .578, .37, .2485, and 64 similar and conterminous. 

3. Make 1.34,45.6, and .341 similar and conterminous. 

4. Make .5674, .34, .247, and 67 similar and conterminous. 

5. Make i.24, .0578, .4, and .4732147 similar and conter- 
minous. 

6. Make .7, .4567, .24, and .346789 similar and conterminous. 

7. Make .8, 36, .4857, .34567, and .2784678943 similar and 
conterminous. 

ADDITION AND SUBTRACTION. 

241. The process of adding and subtracting circulating deci- 
mals depends upon the following properties of repetends : 

I. If two or more repetends are similar and conterminous, their 
denominators will consist of the same number of 9's, with the 
same number of ciphers annexed. Hence, 

II. Similar and conterminous repetends have the same denomi- 
nators and consequently the same fractional unit. 

1. Add .54, 3.24 and, 2.785. 

operation. Analysis. Since fractions can be 

54 54444 added only when they have the same 

Q hi _ Q 91919 fractional unit, we first make the repe- 

. " .^ f tends of the given decimals similar and 

2.785 = 2.78527 conterminous. We then add as in finite 

6.57214 decimals, observing, however, that the 

1 which we carry from the left hand 

column of the repetends, must also be added to the right hand column ; 

for this would be required if the repetends were further expanded 

before adding. 



CIRCULATING DECIMALS. 143 

2. From 7.4 take 2. 7852. 

operation. Analysis. Since one fraction can be subtracted 

7 4.4.4.1 from another only when they have the same frac- 

' . . tional unit, we first make the repetends of the given 

decimals similar and conterminous. We then sub- 
4.6581 tract as in finite decimals ; observing that if both 

repetends were expanded, the next figure in the 
subtrahend would be 8, and the next in the minuend 4 ; and the sub- 
traction in this form would require 1 to be carried to the 2, giving 1 
for the right hand figure in the remainder. 

S54LS. From these principles and illustrations we derive the 
following 

Rule. I. When necessary, make the repetends similar and con- 
terminous. 

II. To add ; — Proceed as in finite decimals, observing to increase 
the sum of the right hand column by as many units as are carried 
from the left hand column of the repetends, 

III. To subtract ; — Proceed as in finite decimals, observing to 
diminish the right hand figure of the remainder by 1, when the 
repetend in the subtrahend is greater than the repetend of the 
minuend. 

IV. Place the points in the result directly under the points above. 

Note. — When the sum or difference is required in the form of a common frac- 
tion, proceed according to the rule, and reduce the result. 

EXAMPLES FOR PRACTICE. 

1. What is the sum of 2.4, .32, .567, 7.056, and 4.37 ? 

Ans. 14.7695877. 

2. What is the sum of .478, .321, .78564, .32, .5, and .4326 ? 

Ans. 2.8961788070698. 

3. From 7854 subtract .59. Ans. .1895258. 

4. From 57.0587 subtract 27.3L Ans. 29.7455. 

5. What is the sum of .5, .32, and 12 ? Ans. 1. 

6. What is the sum of .4387, .863, .2i, and .3554 ? 

7. 'What is the sum of 3.6537, 3.i35, 2.564, and .53 ? 

8. From .432 subtract .25. Ans. 18243. 

9. From 7.24574 subtract 2.634. Ans. 4.6L 



144 



DECIMALS. 



10. From .99 subtract .433. Ans. .556. 

11. What is the sum of 4.638, 8.818, .016, .54, and .45? 

Arts. 13f f 



12. From .4 subtract .23. 

MULTIPLICATION AND DIVISION. 

243. 1. Multiply 2.428571 by .063. 



Arts. 



sV 



OPERATION. 

2.428571 = V 7 
.063 



7 



X 



7_ 

— TlO 

7 17 154 Jfl 9 



Analysis. We first re- 
duce the multiplicand and 
multiplier to their equiva- 
lent fractions, and obtain 
V 7 and TTir ; then K? X T j F 



= Ttt = .154. 



2. Divide .475 by .3753. 

OPERATION. 

.475 =||| 

.3750 = !?^ 

««X 1^ = 1.26 -in.. 



Analysis. The dividend re- 
duced to its equivalent common 
fraction is |££, and the divisor 
reduced to its equivalent com- 
mon fraction is f £f£ ; and |J| 



ttU ==11-1.26. 



24L4:. From these illustrations we have the following 
Rule. Reduce the given numbers to common fractions ; then 
multiply or divide, and reduce the result to a decimal. 



EXAMPLES FOR PRACTICE. 



1. Multiply 3.4 by .72. 

2. Multiply .0432 by 18. 

3. Divide 154 by .2. 

4. Divide 4.5728 by .7. 

5. Multiply 4.37 by .27. 

6. Divide 56.6 by 137. 

7. Divide .42857i by .54. 

8. Multiply .714285 by .27. 

9. Multiply 3.456 by .425. 

10. Divide 9.17054 by 3.36. 

11. Multiply .24 by 57. Arts. 



Ans. 2.472. 

Arts. .7783. 

Ans. .693. 

Ans. 5.8793. 

Ans. 1.182. 

Ans. .41362530. 

Ans. .7857142. 

Ans. 194805. 

Ans. 1.4710037. 

Ans. 2.72637. 

1395775941230486685032. 



UNITED STATES MONEY. ^45 



UNITED STATES MONEY. 

245. By Act of Congress of August 8, 1786, the dollar was 
declared to be the unit of Federal or United States Money ; and 
the subdivisions and multiples of this unit and their denomina- 
tions, as then established, are as shown in the 

TABLE. 

10 mills make 1 cent. 
10 cents " 1 dime. 
10 dimes " 1 dollar. 
10 dollars " 1 eagle. 

24:6. By examining this table we find 

1st. That the denominations increase and decrease in a tenfold 
ratio. 

2d. That the dollar being the unit, dimes, cents and mills are 
respectively tenths, hundredths and thousandths of a dollar. 

3d. That the denominations of United States money increase 
and decrease the same as simple numbers and decimals. 

Hence we conclude that 

I. United States money may be expressed according to the deci- 
mal system of notation. 

II. United States money may be added, subtracted, multiplied 
and divided in the same manner as decimals. 

NOTATION AND NUMERATION. 

247. The character $ before any number indicates that it 

expresses United States money. Thus $75 expresses 75 dollars. 

248. Since the dollar is the unit, and dimes, cents and mills 
are tenths, hundredths and thousandths of a dollar, the deoimal 
point or separatrix must always be placed before dimes. Hence, 
in any number expressing United States money, the first figure at 
the right of the decimal point is dimes, the second figure is cents, 
the third figure is mills, and if there are more than three figures, 
the remainder are ten-thousandths of a dollar. Thus, $8.3125 

13 k 



146 DECIMALS. 

expresses 8 dollars" 3 dimes 1 cent 2 mills and 5 tenths of a mill 
or 5 ten-thousandths of a dollar. 

24:9, The denominations, eagles and dimes, are not regarded 
in business operations, eagles being called tens of dollars and 
dimes tens of cents. Thus $24.19 instead of being read 2 eagles 
4 dollars 1 dime 9 cents, is read 24 dollars 19 cents. Hence, 
practically, the table of United States money is as follows : 

10 mills make 1 cent. 
100 cents " 1 dollar. 

2o0* Since the cents in an expression of United States money 
may be any number from 1 to 99, the first two places at the right 
of the decimal point are always assigned to cents. Hence, when 
the number of cents to be expressed is less than 10, a cipher 
must be written in the place of tenths or dimes. Thus, 7 cents is 
expressed $.07. 

Notes. — 1. The half cent is frequently written as 5 mills and vice versd. 
Thus, $.37* = $.375. 

2. Business men frequently write cents as common fractions of a dollar. 
Thus, $5.19 is also written $5^^, read 5 and T ^ dollars. 

3. In business transactions, when the final result of a computation contains 5 
mills or more, they are called one cent, and when less than 5 they are rejected. 
Thus, $2,198 would be called $2.20, and $1,623 would be called $1.62. 

EXAMPLES FOR PRACTICE. 

1. Write twenty-eight dollars thirty-six cents. 

Arts. $28.36. 

2. Write four dollars seven cents. 

3. Write ten dollars four cents. 

4. Write sixteen dollars four mills. 

5. Write thirty-one and one-half cents. 

6. Write 48 dollars If cents. Ans. $48.01f . 

7. Write 1000 dollars 1 cent 1 mill. 

8. Write 3 eagles 2 dollars 5 dimes 8 cents 4 mills. 

9. Write 6£ cents. 

10. Bead the following numbers : 

$21.18 $10.01 $ .8125 

$164.05 $201,201 $15.08* 

$7.90 $5.37* $96,005 



UNITED STATES MONEY. 147 

REDUCTION. 

251. Since $1 = 100 cents = 1000 mills, it is evident, 

1st, That dollars may be changed or reduced to cents by an- 
nexing two ciphers ; and to mills by annexing three ciphers. 

2d. That cents may be reduced to dollars by pointing off two 
figures from the right ; and mills to dollars by pointing off three 
figures from the right. 

3d. That cents may be reduced to mills by annexing one cipher. 

4th. That mills may be reduced to cents by pointing off one 
figure from the right. 

OPERATIONS IN UNITED STATES MONEY. 

352* Since United States Money may be added, subtracted, 
multiplied and divided in the same manner as decimals, (34U5, 
II), it is evident that no separate rules for these operations are 
required. 

EXAMPLES FOR PRACTICE. 

1. Paid $3475.50 for building a house, $310.20 for painting, 
$1287. 37 } for furniture, and $207.12 J for carpets; how much 
was the cost of the house and furniture ? Ans. $5280.20. 

2. Bought a pair of boots for $4. 62 J, an umbrella for $1.75, a 
pair of gloves for $.87 J, a cravat for $1, and some collars for 
$.62 J; how much was the cost of all my purchases? 

3. Gave $150 for a horse, $175.84 for a carriage, and $62} for 
a harness, and sold the whole for $390.37}; how much did I 
gain? Ans. $12,035. 

4. A man bought a farm for $3800, which was $190.87} less 
than he sold it for ; how much did he sell it for ? 

5. A lady bought a dress for $10 f, a bonnet for $5}, a veil for 
$2|, a pair of gloves for $.87}, and a fan for $-|. She gave the 
shopkeeper a fifty dollar bill ; how much money should he return 
to her? Ans. $29,875. 

6. A farmer sold 150 bushels of oats at $.37} a bushel, and 4 
cords of wood at $3£ a cord. He received in payment 84 pounds of 



248 DECIMALS. 

sugar at 6} cents a pound, 25 pounds of tea at $f a pound, 2 
barrels of flour at $5.87 J, and the remainder in cash; how much 
cash did he receive ? Arts. $39,125. 

7. A speculator bought 264.5 acres of land for $6726. He 
afterward sold 126.25 acres for $31} an acre, and the remainder 
for $33.75 an acre; how much did he gain by the transaction? 

8. A merchant going to New York to purchase goods, had 
$11000. He bought 40 pieces of silk, each piece containing 28 } 
yards, at $.80 a yard; 300 pieces of calicoes, with an average 
length of 29 yards, at 11} cents a yard; 20 pieces of broadcloths, 
each containing 36.25 yards, at $3,875 a yard; 112 pieces of 
sheeting, each containing 30.5 yards, at $.06} a yard. How 
much had he left with which to finish purchasing his stock ? 

Ans. $6064.62}. 

9. If 139 barrels of beef cost $2189.25, how much will 1 
barrel cost? Ans. $15.75. 

10. If 396 pounds of hops cost $44,748, how much are they 
worth per pound ? Ans. $.113. 

11. Bought lOf cords of wood at $4} a cord, and received for 
it 7.74 barrels of flour; how much was the flour worth per barrel? 

12. If a hogshead of wine cost $287.4, how many hogsheads 
can be bought for $4885.80? Ans. 17. 

13. A butcher bought an equal number of calves and sheep for 
$265 ; for the calves he paid $3} a head, and for the sheep $2f- 
a head ; how many did he buy of each kind ? Ans. 40. 

14. If 128 tons of iron cost $9632, how many tons can be 
bought for $1730.75 ? Ans. 23. 

15. If 125 bushels of potatoes cost $41.25, how many barrels, 
each containing 2} bushels, can be bought for $112.20 ? 

16. A grocer on balancing his books at the end of a month, 
found that his purchases amounted to $2475.36, and his sales to 
$1936.40 ; and that the money he now had was but f of what he 
had at the beginning of the month; how much money had he at 
the beginning of the month ? Ans. $1347.40. 

17. A person has an income of $3200 a year, and his expenses 
are $138 a month; how much can he save in 8 years? 



UNITED STATES MONEY. J49 

18. Sold 120 pieces of cloth at $45f a piece, and gained thereby 
$1026 5 how much did it cost by the piece ? Arts. $37.20. 

19. A flour merchant paid $3088.25 for some flour. He sold 
425 barrels at $6i a barrel, and the remainder stood him in $4.50 
a barrel; how many barrels did he purchase ? Arts. 521. 

20. If 36 engineers receive $6315.12 for one month's work, 
how many engineers will $21927.50 pay for one month at the same 
rate? Ans. 125. 

21. A person having $1378.56, wishes to purchase a house 
worth $2538, and still have $750 left with which to purchase fur- 
niture ; how much more money must he have ? Ans. $1909.44. 

22. A mechanic earns on an average $1.87 } a day, and works 22 
days per month. If his necessary expenses are $25f a month, 
how many years will it take him to save $1116, there being 12 
months in a year ? Ans. 6 years, 

23. Bought 27.5 barrels of sugar for $453.75, and sold it at a 
profit of $3.62} a barrel; at what price per barrel was it sold? 

24. A man expended $70.15 in the purchase of rye at $.95 a 
bushel, wheat at $1.37 a bushel, and corn at $.73 a bushel, buying 
the same quantity of each kind ; how many bushels in all did he 
purchase ? Ans. 69 bushels. 

25. A farmer bought a piece of land containing 375} acres, at 
$22J per acre, and sold } of it at a profit of $1032|; at what 
price per acre was the land sold ? Ans. $27.75. 

26 If 3} cords of wood cost $11.37}, how much will 20i cords 
cost? Ans. $65.40|. 

27. If f of a hundred pounds of sugar cost $6|, how much 
can be bought for $46.75, at the same rate? 

Ans. 5.5 hundred pounds. 

28. A man sold a wagon for $62.50, and received in payment 
12} yards of broadcloth at $3 J per yard, and the balance in coffee 
at 12} cents per pound; how many pounds of coffee did he re- 
ceive ? Ans. 175 pounds. 

29. Bought 320 bushels of barley at the rate of 16 bushels for 
$10.04, and sold it at the rate of 20 bushels for $17} ; how much 

as my profit on the transaction ? Ans. $79.20. 

13* 



150 DECIMALS. 

PROBLEMS 
INVOLVING THE RELATION OF PRICE, COST, AND QUANTITY. 

PROBLEM I. 

253. Given, the price and the quantity, to find the cost. 
Analysis. The cost of 3 units must be 3 times the price of 1 unit ; 

of 8 units, 8 times the price of 1 unit ; of f of a unit, f times the price 
of 1 unit, etc. Hence, 

Rule. Multiply the price of one by the quantity. 

PROBLEM II. 

254. Given, the cost and the quantity, to find the price. 

Analysis. By Problem I, the cost is the product of the price mul- 
tiplied by the quantity. Now, having the cost, which is a product, 
and the quantity, which is one of two factors, we have the product 
and one of two factors given, to find the other factor. Hence, 

Rule. Divide the cost by the quantity. 

PROBLEM III. 

255. Given, the price and the cost, to find the quantity. 
Analysis. Reasoning as in Problem II, we find that the cost is 

the product of two factors, and the price is one of the factors, Hence, 

Rule. Divide the cost by the price. 

PROBLEM IV. 

256. Given, the quantity, and the price of 100 or 
1000, to find the cost. 

Analysis. If the price of 100 units be multiplied by the number 
of units in a given quantity, the product will be 100 times the required 
result, because the multiplier used is 100 times the true multiplier. 
For a similar reason, if the price of 1000 units be multiplied by the 
number of units in a given quantity, the product will be 1000 times 
the required result. These errors can be corrected in two ways, 
1st. By dividing the product by 100 or 1000, as the case may be; or, 
2d. By reducing the given quantity to hundreds and decimals of a 
hundred, or to thousands and decimals of a thousand. Hence, 



PROBLEMS IN UNITED STATES MONEY. \tft 

Rule. Multiply the price by the quantity reduced to hundreds 

and decimals of a hundred^ or to thousands and decimals of a 

thousand. 

Note. — In business transactions the Roman numerals C and M are com- 
monly used to indicate hundreds and thousands, where the price is by the 100 
or 1000. 

PROBLEM V. 

257. To find the cost of articles sold by the ton of 
2000 pounds. 

Analysis. If the price of 1 ton or 2000 pounds be divided by 2, 
the quotient will be the price of J ton or 1000 pounds. We then have 
the quantity and the price of 1000 to find the cost. Hence, 

Rule. Divide the price of 1 ton by 2, and multiply the quo- 
tient by the number of pounds expressed as thousandths. 

EXAMPLES IN THE PRECEDING PROBLEMS. 

1. What cost 187 barrels of salt, at $1.32 a barrel? 

Ans. $246.84. 

2. What cost 5 firkins of butter, each containing 70| pounds, 
at $ T 3 g a pound ? Ans. $66.09f . 

3. If the board of a family be $501,872 for 1 year, how much 
is it per day ? Ans. $1.37£. 

4. At $.10| a dozen, how many dozen of eggs can be bought 
for $18.48 ? Ans. 176. 

5. What is the value of 140 sacks guano, each sack containing 
162* pounds, at $17f a ton? Ans. $201.906|. 

6. What will be the cost oi 3240 peach trees at $16| per hun- 
dred? Ans. $534.60. 

7. At $66.44 a ton, what will be the cost of 842| tons of rail- 
road iron? Ans. $55992.31. 

8. A gentleman purchased a farm of 325.5 acres for $10660| ; 
how much did it cost per acre? Ans. $32.75. 

9. What will be the cost of 840 feet of plank at $1.94 per C; 
and 1262 pickets at $12 £ per M ? Ans. $32,071. 

10. At $1^ a bushel, how many bushels of wheat can be bought 
for $37.68| ? Ans. 25£ bushels. 



J52 DECIMALS. 

. 11. What will be the cost of 2172 pounds of plaster, at $3,875 
a ton? Am. $4.208f 

12. What cost | of 456 bushels of potatoes at $.37^ a bushel ? 

13. If 32£ barrels of apples cost $81.25, what is the price per 
barrel? Am. $2.50. 

14. What must be paid for 24240 feet of timber worth $9.37 } 
perM.? Am. $227}. 

15. At $5| an acre, how many acres of land can be bought for 
$4234.37}? Am. 752|. 

16. How much must be paid for 972 feet of boards at $20.25 
per M, 1575 feet of scantling at $2.87} per C, and 8756 feet of 
lath at $7 } per M ? Am. $130,634*. 

«•■ 17. What is the value of 1046 pounds of beef at $4f per hun- 
dred pounds? Am. $48.37f. 

18. What is the value of 5840 pounds of anthracite coal at 
$4.7 a ton, and 4376 pounds of shamokin coal at $5.25 a ton ? 

19. At $2.50 a yard, how much cloth can be purchased for $2 ? 

20. What is the value of 3700 cedar rails at $5f per C ? 

21. How much is the freight on 3840 pounds from New York 
to Baltimore, at $.96 per 100 pounds ? Am. $36,864. 

22. What is the value of 9 pieces of broadcloth, each piece 
containing 27f yards, worth $21 a yard ? Am. $715.87 J. 

23. At $.42 a pound, how many pounds of wool may be bought 
for $80,745? Am. 192}. 

24. What will be the cost of 327 feet of boards at $15} per 
M; 672 feet of siding at $1.62} per C, and 1108 bricks at $4} 
per M? Am. $20.69f. 

25 At $i per yard, how many yards of silk may be bought for 
$15|? Am. 18. 

26. How much must be paid for the transportation of 18962 
pounds of pork from Cincinnati to New York, at $10 a ton? 

27. If 15} yards of silk cost $27.9, what is the price per yard? 

28. What cost 27860 railroad ties at $125.38 per thousand? 

29. If .7 of a ton of hay cost $13J, what is the price of 1 ton ? 

30. What is the value of 720 pounds of hay at $12.75 a ton, 
and 912 pounds of mill feed at $15} a ton? Am. $11,658. 






ACCOUNTS AND BILLS. 



153 



LEDGER ACCOUNTS. 
358. A Ledger is the principal book of accounts kept by mer- 
chants and accountants. Into it are brought in summary form 
the accounts from the journal or day-book. The items often form 
long columns, and accountants in adding sometimes add more than 
one column at a single operation, (68). 



a-) 



(2-) 



(3.) 



(40 



& 42.17 


$ 506.76 


•2371.67 


$14763.84 


36.24 


194.32 


4571.84 


33276.90 


18.42 


427.90 


1690.50 


47061.39 


10.71 


173.26 


2037.69 


18242.76 


194.30 


71.32 


5094.46 


37364.96 


347.16 


39.46 


876.54 


8410.31 


40.00 


152.60 


679.81 


5724.27 


12.94 


271.78 


155.48 


56317.66 


86.73 


320.00 


4930.71 


81742.73 


271.19 


709.08 


3104.13 


22431.27 


103.07 


48.50 


1987.67 


40163.55 


500.50 


63.41 


5142.84 


32189.60 


7.59 


56.00 


276.30 


7063.21 


11.44 


410.10 


522.71 


3451.09 


81.92 


372.22 


3114.60 


9200.00 


110.10 


137.89 


1776.82 


1807.36 


107.09 


276.44 


7152.91 


56768.72 


207.16 


18.19 


9328.42 


63024.27 


97.20 


27.96 


472.19 


3618045 


21.77 


157.16 


321.42 


90807.08 


150.15 


94.57 


2423.79 


28763.81 


427.26 


177.66 


1600.81 


37196.75 


316.42 


327.40 


5976.27 


4230 61 


114.64 


1132.16 


4318.19 


3719.84 


81.13 


876.57 


682,45 


1367.92 


37.50 


179.84 


3174.96 


8756.47 



ACCOUNTS AND BILLS. 

250* A Debtor, in business transactions, is a purchaser, or a 
person who receives money, goods, or services from another; and 

*260. A Creditor is a seller, or a person who parts with 
money, goods, or services to another. 



154 DECIMALS. 

261. An Account is a registry of debts and credits. 

Notes.— 1. An account should always contain the names of both the parties to 
the transaction, the price or value of each item or article, and the date of the 
transaction. 

2. Accounts may have only one side, which may be either debt or credit; or it 
may have two sides, debt and credit. 

262. The Balance of an Account is the difference between 
the amount of the debit and credit sides. If the account have 
only one side, the balance is the amount of that side. 

263. An Account Current is a full copy of an account, 
giving each item of both debit and credit sides to date. 

SS® 4. A Bill, in business transactions, is an account of money 
paid, of goods sold or delivered, or of services rendered, with the 
price or value annexed to each item. 

265. The Footing of a Bill is the total amount or cost of all 
the items. 

Note. — A bill of goods bought or sold, or of services received or rendered at 
a single transaction, and containing only one date, is often called a Bill of Par- 
cels ; and an account current having only one side is sometimes called a Bill 
of Items, 

260. In accounts and bills the following abbreviations are in 
general use : 

Br. for debit or debtor ; 
Or. for credit or creditor ; 
a | c . or acc't for account; 

@ for at or by; when this abbreviation is used it is always 
followed by the price of a unit. Thus, 3 yd. cloth @ $1.25, sig- 
nifies 3 yards of cloth at $1.25 per yard; J lb. tea @ $.75, signi- 
fies i pound of tea at $.75 per pound. 

267. When an account current or a bill is settled or paid, 
the fact should be entered on the same and signed by the creditor, 
or by the person acting for him. The a | c . or bill is then said to 
be receipted. Accounts and bills may be settled, balanced and 
receipted by the parties to the same, or by agents, clerks or attor- 
neys authorized to transact business for the parties. 



ACCOUNTS AND BILLS. 



155 



EXAMPLES FOR PRACTICE. 

Required, the footings and balances of the following bills and 
accounts. 

a) 

Bill : receipted by clerk or agent. 

New York, July 10, 1860. 
Mr. John C. Smith, 







Bo 


't of Hill, Groves & Co., 


10 yd. 
16 " 


Cassimere, 
Blk. Silk, 




@ $2.85 
" 1.121 




36 " 
42 « 
12 " 


Ticking, 
Bid. Shirting, 
Pressed Flannel, 




" .28 
" .161 
« .40 




241" 


Scotch Plaid Prints, 


« .56 




Pec'd Pc 


lyment, 




$82.03 
Hill, Groves & Co., 








By J. 


W. Hopkins. 



(2.) 
Bill: receipted by the selling party. 

Chicago, Sept. 20, 1861. 
Chase & Kennard, 

Bdtof McDougal, Fenton & Co., 

125 pr. Boys' Thick Boots, @ $1.25 



Calf " " 

Kip " " 

Brogans, a 

Women's Fox'd Gaiters, " 

Opera Boots, " 

" Enameled " 

8 cases Men's Calf Boots, 

3 " Congress Pump Boots, 

1 " Drill, 958 yd., 

40 gross Silk Buttons, 



275 


ii 


a 


180 


u 


u 


210 


u 


a 


80 


u 


Wo 


95 


u 




175 


a 





1.75 
1.121 

.871 

.84 

.90 

1.06 

30.50 

35.75 

.101 



Hec'd Payment, 



$1828.79 
McDougal, Fenton & Co. 



>6 DECIMALS. 






(3.) 






Bill: settled 


by note. 






New York, May 4, 1860. 


Smith & Perkins, 










Bo't of Kent, Lowber & Co., 


40 chests Green Tea, 




@ $27.50 


25 " Black, 




u 


19.20 


16 " Imperial, 




a 


48.10 


12 sacks Java Coffee, 




tt 


17.75 


20 bbl. Coffee Sugar 


,(A) 


tt 


26.30 


15 " Crushed, 




tt 


31.85 


36 boxes Lemons, 




tt 


3.875 


42 " Oranges, 




a 


4.12J 


25 " Raisins, 




a 


2.90 




$2961.60 


Rec'd Payment^ by note at 6 mo. 










Kent, Lowber & Co. 



. 



(40 

Bill: paid by draft, and receipted by Clerk. 

New Orleans, April 28, 1861. 
James Carlton & Co. 





Bo't of Willard & Hale. 


150 bbl. Canada Flour, 


@ $6.25 


275 " Genesee, 




" 7.16 


170 " Philadelphia, 


" 5.87J 


326 bu. Wheat, 




« 1.62* 


214 " Corn, 




« .82 


300 " Barley, 




" .91 


500 " Eye, 




" 1.06 




$5413.48 


iment, by Draft on 


N. Y. 






R. S. Clarke, 






For Willard & Hale. 



ACCOUNTS AND BILLS. 



157 



(5.) 

Account Current ; not balanced or settled. 

Philadelphia, Nov. 1, 1860. 
Mr. James Cornwall, 

To Dodge & Son, Dr. 

April 15, To 24 tons Swedes Iron, @ $64.30 $ 

6 " « 15 cwt. Eng. Blister Steel, " 10.25 



June 21, 


" 7 doz. Hoes, (Trowel Steel) « 


7.78 


Aug. 10, 


" 25 " Buckeye Plows 


u 


8.45 


Oct. 3, 


" 14 Cross-cut Saws, 


u 


16.12J 


u u 


" 27 cwt. Bar Lead, 


u 


5.90 


it u 


" 1840 lbs. Chain, 

Cr. 


« 


.09J 




i 


May 25, 


By 20 M. Boards, @ $17.60 




July 14, 


« 50 M. Shingles, " 


3.121 




U (t 


" 42 M. Plank, " 


9.37* 




Sept. 5, 


" Draft on New York, 




$1000 


« 12, 


" 75 C. Timber, @ 


3.10 




« « 


" 36 C. Scantling, « 


.87J 






$ 



Bal. Due Dodge & Son, $256.51 



(6.) 



Account Current, another form ; balanced by note. 
Wm. Richmond & Co. in a|c. current with Wood & Powell. 



Dr. 



Or. 



1850 

July 

Aug. 

« 

Oct. 



Dec. 



To 896 pounds butter, $.23 

« 872 " cheese, .09^ 

" 481% " lard, .11^ 

" 509% " tallow, .13'^ 

" 81 dozen eggs, .16^ 3 

" 15 barrels salt, 1.40 
« 41 hams, 963% pounds, .12% 







1860 








Nov. 








a 


24 






Dec. 


1 






u 


22 






1861 








Jan. 


2 


565 


25 







By 61 barrels apples, $2.25 
" 70 bushels turnips, .22 
" 56 " dried apples, .87^ 
" 31 drums figs, .68^; 

" Note at 3 mo. to Bal. 



Boston, Jan. 1, 1861. 



Wood & Powell. 



158 DECIMALS. 

PROMISCUOUS EXAMPLES. 

1. What cost 12| cords of wood® $4,871? Am. $61.54+. 

2. At $.37} per busUel, how many barrels of potatoes, each 
containing 2} bushels, can be purchased for $33.75 ? Arts. 36. 

3. If 36 boxes of raisins, each containing 36 pounds, can be 
bought for $97.20, what is the price per pound ? Arts. $.075. 

4. If .625 of a barrel of flour be worth $5.35, what is a barrel 
worth ? Am. $8.56. 

5. What is the difference between | of a hundredth, and | of 
a tenth ? Am. .025. 

6. What is the product of 814^ x 26|| correct to 7 decimal 
places ? 

7. A drover bought 5 head of cattle @ $75, and 12 head @ 
$68 ; at what price per head must he sell them to gain $118 on 
the whole ? Am. $77. 

8. If 1 pound of tea be worth $.62}, what is .8 of a pound 
worth? Arts. $.5. 

9. A person having $27.96, was desirous of purchasing an 
equal number of pounds of tea, coffee, and sugar* the tea @ 
$.87}, the coffee @ $.18f, and the sugar @ $.10J. How many 
pounds of each could he buy? Am. 24. 

10. If a man travel 13543.47 miles in 3651 days, how far does 
he travel in I of a day ? Arts. 32.445 miles. 

11. Bought 100 barrels of flour @ $5.12}, and 250 bushels of 
wheat @ $1.06J. Having sold 75 barrels of the flour @ $6}, 
and all the wheat @ $1|, at what price per barrel must the re- 
mainder of the flour be sold, to gain $221.87} on the whole invest- 
ment? Am. $6.75. 

12. If 114.458 acres of land produce 4580.289 bushels of pota- 
toes, how many acres will be required to produce 120.06 bushels ? 

Arts. 3. 

13. Divide .0172 1| by .03^, and obtain a quotient true to 4 
decimal places. Am. .5625db. 

14. Divide 13.5 by 21 hundredths. Am. 600. 

15. A man agreed to build 59.5 rods of wall; having built 8.5 



PROMISCUOUS EXAMPLES. 159 

rods in 5 days, how many days will be required to finish the wall 
at the same rate ? Arts. 30 days. 

16. A farmer exchanged 28} bushels of oats worth $.37} per 
bushel, and 453 pounds of mill feed worth $.75 per hundred, for 
12520 pounds of plaster; how much was the plaster worth per 
ton ? Ans. $2.25. 

17. A farmer sold to a merchant 3 loads of hay weighing re- 
spectively 1826, 1478, and 1921 pounds, at $8.80 per ton, and 
281 pounds of pork at $5.25 per hundred. He received in exchange 
31 yards of sheeting @ $.09, 6} yards of cloth @ $4.50, and the 
balance in money; how much money did he receive? 

18. If 35 yards of cloth cost $122.50, what will be the cost of 
29 yards? Ans. $101.50. 

19. A speculator bought 1200 bushels of corn @ $.56i. He 
sold 375} bushels @ $.6. At what price must he sell the re- 
mainder, to gain $168,675 on the whole ? 

20. If a load of plaster weighing 1680 pounds cost $2,856, how 
much will a ton of 2000 pounds cost? Ans. $3.40. 

21. If .125 of an acre of land is worth $15|, how much are 
25.42 acres worth ? 

22. A farmer had 150 acres of land, which he could have sold 
at one time for $100 an acre, and thereby have gained $3900 ; but 
after keeping it for some time he was obliged to sell it at a loss 
of $2250. How much an acre did the land cost him, and how 
much an acre did he sell it for ? 

23. A lumber dealer bought 212500 feet of lumber at $14,375 
per M, and retailed it out at $1.75 per C ; how much was his 
whole gain ? 

24. If 10 acres of land can be bought for $545, how many 
acres can be bought for $17712.50 ? Ans. 325. 

25. How much is the half of the fourth part of 7 times 224.56 ? 

Ans. 196.49. 

26. Sold 10450 feet of timber for $169.8125, and gained 
thereby $39.18| ; how much did it cost per C ? Ans. $1.25. 

27. If $6,975 be paid for .93 of a hundred pounds of pork, 
how much will 1 hundred pounds cost ? 



160 



DECIMALS. 



28. Three hundred seventy-five thousandths of a lot of dry 
goods, valued at $4000, was destroyed by fire ; how much would 
a firm lose who owned .12 of the entire lot { Arts. $180. 

29. Reduce ( jf -*- of) X I of i to a decimal. Arts. .15. 

30. If 7.5 tons of hay are worth 375 bushels of potatoes, and 
1 bushel of potatoes is worth $.33|, how much is 1 ton of hay 
worth? Ans. $16.66§. 

31. A person invested a certain sum of money in trade; at the 
end of 5 years he had gained a sum equal to 84 hundredths of it, 
and in 5 years more he had doubled this entire amount. How 
many times the sum first invested had he at the end of the 10 
years? Arts. 3.68 times. 

32. A miller paid $54 for grain, T 3 ^ of it being barley at $.62 J 
per bushel, and f of it wheat at $1.87 J per bushel; the balance 
of the money, he expended for oats at $.37} per bushel. How 
many bushels of grain did he purchase ? Ans. 40. 

33. A merchant tailor bought 27 pieces of broadcloth, each 
piece containing 19£ yards, at $4.31 J a yard; and sold it so as to 
gain $381.87}, after deducting $9.62} for freight. How much 
was the cloth sold for per yard ? Ans. $5.06 J. 

34. Bought 1356 bushels of wheat @ $1.18f , and 736 bushels 
of oats @ $.41 ; I had 870 bushels of the wheat floured, and dis- 
posed of it at a profit of $235.87}, and I sold 528 bushels. of the 
oats at a loss of $13.62}. I afterward sold the remainder of the 
wheat at $1.12} per bushel, and the remainder of the oats at $.31 
per bushel ; did I gain or lose, and how much ? 

Ans. I gained $189.79}. 

35. The sum of two fractions is }§ §, and their difference is 
i^| ; what are the fractions ? 

36. A manufacturer carried on business for 3 years. The first 
year he gained a sum equal to | of his original capital ; the second 
year he lost \ of what he had at the end of the first year ; the 
third year he gained | of what he had at the end of the second 
year, and he then had $28585.70. How much had he gained in 
the 3 years? Ans. $10595.60. 



CONTINUED FRACTIONS. 161 

CONTINUED FRACTIONS. 

268. If we take any fraction in its lowest terms, as |f , and 
divide both terms by the numerator, we shall obtain a complex 
fraction, thus : 

13_1 

54 ~~ 4 + 2^ 
13 
Reducing ^, the fractional part of the denominator, in the same 
manner, we have, 

13_1 

54 ~~ 4 + 1 

6 + l_ 
2 
Expressions in this form are called continued fractions. Hence, 
369. A Continued Fraction is a fraction whose numerator is 
1, and whose denominator is a whole number plus a fraction 
whose numerator is also 1, and whose denominator is a similar 
fraction, and so on. 

370. . The Terms of a continued fraction are the several sim- 
ple fractions which form the parts of the continued fraction. 
Thus, the terms of the continued fraction given above are, i, i, 

and J. 

CASE I. 

271* To reduce any fraction to a continued fraction. 

1. Reduce ||| to a continued fraction. 

operation. Analysis. We divide the denominator, 

109 1 339, by the numerator, 109, and obtain 3 

339 3+1 for the denominator of the first term of 

9 i i the continued fraction. Then in the same 
-To manner we divide the last divisor, 109, by 
the remainder, 12, and obtain 9 for the de- 
nominator of the second term of the continued fraction. In like man- 
ner we obtain 12 for the denominator of the final term. Hence the 
following 

Rule. I. Divide the greater term by the less, and the last 
divisor by the last remainder, and so on, till there is no remainder. 
14* L 



2g2 CONTINUED FRACTIONS. 

II. Write 1 for the numerator of each term of the continued 
fraction, and the quotients in their order for the respective denom- 
inators. 



EXAMPLES FOR PRACTICE 

1. Reduce T 2 7 8 ^ to a continued fraction. 



Ans. —-— : 



6 + L_ 

4 + l_ 
2 + l_ 

9* 

2. Reduce |§|f to a continued fraction. 

3. Reduce §f§f ^jf to a continued fraction. 

4. Reduce T 2 ^ 9 T to a continued fraction. 

CASE II. 

272. To find the several approximate values of a 
continued fraction. 

An Approximate Value of a continued fraction is the simple 
fraction obtained by reducing one, two, three, or more terms of the 
continued fraction. 

273. 1. Reduce -^^ to a continued fraction, and find its 
approximate values. 

OPERATION. 

38 1 

=-T£ = n~r> tne continued fraction. 
163 4+- 1 

3+T 
3 + 1 

2 + 1 
5 

= i 1st approx. value. 

1 4X3 + 1 13' - 

4 + 1 

3 

1__ = 1 3 X2 + 1 - 3X2 + 1 ^- 7.3d « 

4 + 1 4 + 2 (4 X3 + i) X 2 + 4 13 X 2 + 4 30 

3+1 3X2+1 

L_ 2 = _lX_ 5 J_ 3 = ^,4th« « 

4 + 30X5+13 163' 

3~+T_ 

2 + 1 

5 



CONTINUED FRACTIONS. 163 

Analysis. We take |, the first term of the continued fraction, 

for the first approximate value. Eeducing the complex fraction 
formed by the first two terms of the continued fraction, we have T \ 
for the second approximate value. In like manner, reducing the first 
three terms, we have ^ for the third approximate value. By exam- 
ining this last process, we perceive that the third approximate value, 
s 7 o , is obtained by multiplying the terms of the preceding approxima- 
tion, T %, by the denominator of the third term of the continued frac- 
tion, 2, and adding the corresponding terms of the first approximate 
value. Taking advantage of this principle, we multiply the terms of 
So by the 4th denominator, 5, in the continued fraction, and adding 
the corresponding terms of j\, obtain T ^, the 4th approximate value, 
which is the same as the original fraction. Hence the following 

Rule. I. For the first approximate value, take the first term 
of the continued fraction. 

II. For the second approximate value, reduce the complex- frac- 
tion formed by the first two terms of the continued fraction. 

III. For each succeeding approximate value, multiply both nu- 
merator and denominator of the last preceding approximatien by 
the next denominator in the continued fraction, and add to the cor- 
responding products respectively the numerator and denominator of 
the preceding approximation. 

Notes. — 1. When the given fraction is improper, invert it, and reduce this 
result to a continued fraction,* then invert the approximate values obtained 
therefrom. 

2. In a series of approximate values, the 1st, 3d, 5th, etc., are less than the 
given fraction • and the 2d, 4th, 6th, etc., are greater than the given fraction. 

EXAMPLES FOR PRACTICE. 

1. Find the approximate values of T 6 ^. 



2. Find the approximate values of 3^. 



Ant 1 3 16 67 



An* 1 4 5 3 9_ J33 
J±llb. 4 , jyj, 2 T , y 64 , 34 ^. 



3. What are the first three approximate values of /oW? ? 



An* 1 5 - 2 - 1 - 
j*nb. 7, 36 , j 51 . 



4. What are the first five approximate values of § || ? 

A. no 1 4 9 4 4_9 

5. Reduce || to the form of a continued fraction, and find the 
value of each approximating fraction. 



Ig4 COMPOUND NUMBERS. 



COMPOUND NUMBEKS. 

374. A Compound Number is a concrete number expressed 
in two or more denominations, (10). 

275* A Denominate Fraction is a concrete fraction whose 
integral unit is one of a denomination of some compound number. 
Thus, | of a day is a denominate fraction, the integral unit being 
one day; so are § of a bushel, § of a mile, etc., denominate trac- 
tions. 

276. In simple numbers and decimals the scale is uniform, 
and the law of increase and decrease is by 10. But in compound 
numbers the scale of increase and decrease from one denomination 
to another is varying, as will be seen in the Tables. 

MEASUKES. 

277. Measure is that by which extent, dimension, capacity 
or amount is ascertained, determined according to some fixed 
standard. 

Note. — The process by which the extent, dimension, capacity, or amount is 
ascertained, is called Measuring ; and consists in comparing the thing to be 
measured with some conventional standard. 

Measures are of seven kinds : 

1. Length. 4. Weight, or Force of Gravity. 

2. Surface or Area. 5. Time. 

3. Solidity or Capacity. 6. Angles. 

7. Money or Value. 
The first three kinds may be properly divided into two classes — 
Measures of Extension, and Measures of Capacity. 

MEASURES OF EXTENSION. 

278. Extension has three dimensions — length, breadth, and 
thickness. 

A Line has only one dimension — length. 

A Surface or Area has two dimensions — length and breadth. 



MEASURES OF EXTENSION. 165 

A Solid or Body has three dimensions — length, breadth, and 
thickness. 

I. Linear Measure. 

279. Linear Measure, also called Long Measure, is used in 
measuring lines or distances. 

The unit of linear measure is the yard, and the table is made 
up of the divisors, (feet and inches,) and the multiples, (rods, 
furlongs, and miles,) of this unit. 

TABLE. 

12 inches (in.) make 1 foot, ft. 

3 feet " 1 yard, yd. 

5 J yards, or 16} feet, " 1 rod, rd. 

40 rods " 1 furlong, fur. 

8 furlongs, or 320 rods, " 1 statute mile,.. mi. 

UNIT EQUIVALENTS. 

ft. in. 

yd. 1 = 12 

rd. 1 = 3 = 36 

fur. 1 = 5J = 16} = 198 

mi 1 = 40 = 220 = 660 = 7920 

t = 8 = 320 = 1760 = 5280 = 63360 
Scale — ascending, 12, 3, 5J, 40, 8; descending, 8, 40, 5J, 3, 12. i 
The following denominations are also in use : — 

. , ; t . rtW f used by shoemakers in measuring 

3 barleycorns make 1 inch, j the ^ gth of ^ foot> « 

. , . n , -, f used in measuring the height of 

4 inches 1 hand, j horges directly ^ ^ for | feet> 

9 " "1 span. 

21.888 " " 1 sacred cubit. 

3 feet " 1 pace. 

6 " "1 fathom, used in measuring depths at sea. 

1.16 statute miles « 1 geographic mile, { ^es™™" * 

3 geographic " " 1 league. 

60 " " or l i ^ f ^ latitude on a meridian or of 

69.16 statute " " J ae g ree j longitude on the equator. 

360 degrees " the circumference of the earth. 

Notes. — 1. For the purpose of measuring cloth and other goods sold by the 
yard, the yard is divided into halves, fourths, eighths, and sixteenths. The old 
table of cloth measure is practically obsolete. 

2. A span is the distance that can be reached, spanned, or measured between 
the end of the middle finger and the end of the thumb. Among sailors 8 spans 
are equal to 1 fathom. 

3. The geographic mile is Vo °f 3^0 or 2 f^u of tne distance round the center 
of the earth. It is a small fraction more than 1.15 statute miles. 



166 



COMPOUND NUMBERS. 



4. The length of a degree of latitude varies, being 68.72 miles at the equator, 
68.9 to 69.05 miles in middle latitudes, and 69.30 to 69.34 miles in the polar 
regions. The mean or average length, as stated in the table, is the standard 
recently adopted by the U. S. Coast Survey. A degree of longitude is greatest 
at the equator, where it is 69.16 miles, and it gradually decreases toward the 
poles, where it is 0. 

surveyors' linear measure. 

380. A Gunter's Chain, used by land surveyors, is 4 rods 
or 66 feet long, and consists of 100 links. 

The unit is the chain, and the table is made up of divisors and 
multiples of this unit. 

TABLE. 

7.92 inches (in.) make 1 link, 1. 

25 links " 1 rod, rd. 

4 rods, or 66 feet, " 1 chain, . . . ch. 

80 chains " 1 mile, .... mi. 



UNIT EQUIVALENTS. 













1. 




in. 








rd. 




1 


= 


7.92 




ch. 




1 


= 


25 


= 


198 


mi. 


1 


SB 


4 


= 


100 


BBS 


792 


1 = 


80 


= 


320 


= 


8000 


= 


63360 



Scale — ascending, 7.92, 25, 4, 80; descending, 80, 4, 25, 7.92. 

Note. — The denomination, rods, is seldom used in chain measure, distances 
being taken in chains and links. 

II. Square Measure. 

281. A Square is a figure having four equal sides and four 
equal corners or right angles. 

282. Area or Superficies is the space or surface included 
within any given lines : as, the area of a square, of a field, of a 
board, etc. 

1 square yard is a figure having four 
sides of 1 yard or 3 feet each, as shown 
in the diagram. Its contents are 3x3 
= 9 square feet. Hence, 

The contents or area of a square, or 
of any other figure having a uniform 
length and a uniform breadth, is found 
by multiplying the length by the breadth. 



1 yd. = 3 ft. 





















MEASURES OF EXTENSION. J67 

Thus, a square foot is 12 inches long and 12 inches wide, and the 
contents are 12 X 12 = 144 square inches. A board 20 feet long and 
10 feet wide, is a rectangle, containing 20 X 10 = 200 square feet. 

The measurements for computing area or surface are always 
taken in the denominations of linear measure. 

283. Square Measure is used in computing areas or sur- 
faces ) as of land, boards, painting, plastering, paving, etc. 

The unit is the area of a square whose side is the unit of 
length. Thus, the unit of square feet is 1 foot square ; of square 
yards, 1 yard square, etc. 

TABLE. 
144 square inches (sq. in.) make 1 square foot, sq. ft. 



9 " feet 

30} " yards 
40 " rods 

4 roods 

640 acres 



1 " yard,...sq. yd. 

1 " rod, . . . . sq. rd. 

1 rood, R. 

1 acre, A. 

1 square mile,. . . sq. mi. 



UNIT EQUIVALENTS. 

sq. ft. sq. in. 

sq.yd. 1 = 144 

sq.rd. 1 = 9 = 1296 

r. 1 = 30}= 272}= 39204 

A . 1 = 40 = 1210 = 10890 = 1568160 

sq.mi. 1 = 4 = 160 = 4840 = 43560 = 6272640 

1 = 640 = 2560 = 102400 = 3097600 == 27878400 = 4014489600 

Scale — ascending, 144, 9, 30}, 40, 4, 640 ; descending, 640, 4, 40, 

30}, 9, 144. 

Artificers estimate their work as follows : 

By the square foot : glazing and stone-cutting. 

By the square yard : painting, plastering, paving, ceiling, and 
paper-hanging. 

By the square of 100 feet: flooring, partitioning, roofing, 
slating, and tiling. 

Brick-laying is estimated by the thousand bricks, by the square 

\ yard, and by the square of 100 feet. 

Notes. — 1. In estimating the painting of moldings, cornices, etc., the mea- 
suring-line is carried into all the moldings and cornices. 

2. In estimating brick-laying by either the square yard or the square of 100 
I feet, the work is understood to be 12 inches or 1£ bricks thick. 

3. A thousand shingles are estimated to cover 1 square, being laid 5 inches to 
the weather. 



168 



COMPOUND NUMBERS. 



SURVEYORS' SQUARE MEASURE. 

384. This measure is used by surveyors in computing the 
area or contents of land. 

TABLE. 

make 



625 square links (sq. 1.) 

16 poles 

10 square chains 
640 acres 

36 square miles (6 miles square) 



1 pole, . P. 

1 square chain, . sq. ch. 

1 acre, A. 

1 square mile, . sq. mi. 
1 township, Tp. 



Tp. 
1 



sq. mi. 
36 = 





UNIT EQUIVALENTS 


p. 




sq. 1. 




sq. ch. 


1 


= 


625 


A. 


1 = 


16 


= 


1000 


1 


= 10 = 


160 


= 


10000 


640 


= 6400 = 


102400 


: — 


64000000 


040 


= 230400 = 


3686400 


= 


2304000000 



descending, 36, 640, 10, 



Scale — ascending, 625, 16, 10, 640, 36 ; 
16, 625. 

Notes. — 1. A square mile of land is also called a section. 

2. Canal and railroad engineers commonly use an engineers' chain, which con- 
sists of 100 links, each 1 foot long. 

3. The contents of land are commonly estimated in square miles, acres, and 
hundredths; the denomination, rood, is rapidly going into disuse. 

III. Cubic Measure. 

285. A Cube is a solid, or body, having six equal square 
sides or faces. 

286. Solidity is the matter or space contained within the 
bounding surfaces of a solid. 

The measurements for computing solidity are always taken in 
the denominations of linear measure. 

If each side of a cube be 1 yard, or 3 
feet, 1 foot in thickness of this cube will 
contain 3x3x1 = 9 cubic feet ; and the 
whole cube will contain 3 x 3 X 3 = 27 
cubic feet. 

A solid, or body, may have the three 
dimensions all alike or all different. A 
body 4 ft. long, 3 ft. wide, and 2 ft. thick 
contains 4 x 3 x 2 = 24 cubic or solid 
feet. Hence we sec that 




a ft. = 1 yd. 



MEASURES OF EXTENSION. \QQ 

TJie cubic or solid contents of a body are found by multiplying 
the length, breadth, and thickness together, 

987. Cubic Measure, also called Solid Measure, is used in 
computing the contents of solids, or bodies; as timber, wood 
stone, etc. 

The unit is the solidity of a cube whose side is the unit of 
length. Thus, the unit of cubic feet is a cube which measures 1 
foot on each side ; the unit of cubic yards is 1 cubic yard, etc. 



TABLE. 

1728 cubic inches (cu. in.) make 1 cubic foot cu. ft. 

27 cubic feet 

40 cubic feet of round timber, or ) 
50 "■ " hewn " j 

16 cubic feet 

8 cord feet, or \ 

128 cubic feet J 



24| cubic feet 



1 cubic yard cu. yd. 

1 ton or load T. 

1 cord foot ccL ft. 

1 cord of wood ..... Cd. 

-j f perch of stone ) p , 
( or masonry, J 



Scale — ascending, 1728, 27. The other numbers are not in a 
regular scale, but are merely so many times 1 foot. The unit equiva- 
lents, being fractional, are consequently omitted. 

Notes. — 1. A cubic yard of earth is called a load. 

2. Railroad and transportation companies estimate light freight by the space 
it occupies in cubic feet, and heavy freight by weight. 

3. A pile, of wood 8 feet long, 4 feet wide, and 4 feet high, contains 1 cord; 
and a cord foot is 1 foot in length of such a pile. 

4. A perch of stone or of masonry is 16£ feet long, 1£ feet wide, and 1 foot 
high. 

5. Joiners, bricklayers, and masons, make no allowance for windows, doors, 
etc. Bricklayers and masons, in estimating their work by cubic measure, make 
no allowance for the corners of the walls of houses, cellars, etc., but estimate 
their work by the girt, that is, the entire length of the wall on the outside. 

6. Engineers, in making estimates for excavations and embankments, take the 
dimensions with a line or measure divided into feet and decimals of a foot. The 
computations are made in feet and decimals, and the results are reduced to cubic 
yards. In civil engineering, the cubic yard is the unit to which estimates for 
excavations and embankments are finally reduced. 

7. In scaling or measuring timber for shipping or freighting, £ of the solid 
contents of round timber is deducted for waste in hewing or sawing. Thus, a 
log that will make 40 feet of hewn or sawed timber, actually contains 50 cubic 
feet by measurement; but its market value is only equal to 40 cubic feet of 
hewn or sawed timber. Hence, the cubic contents of 40 feet of round and 50 
feet of hewn timber, as estimated for market, are identical. 

15 



170 



COMPOUND NUMBERS. 



MEASURES OP CAPACITY. 



388. Capacity signifies extent of room or space. 

289. Measures of capacity are all cubic measures, solidity and 
capacity being referred to different units, as will be seen by com- 
paring the tables. 

Measures of capacity may be properly subdivided into two 
classes, Measures of Liquids and Measures of Dry Substances. 

I. Liquid Measure. 

290. Liquid Measure, also called Wine Measure, is used in 
measuring liquids ; as liquors, molasses, water, etc. 

The unit is the gallon, and the table is made up of its divisors 
and multiples. 

table. 

4 gills (gi.) make 1 pint, pt. 

2 pints " 1 quart, qt. 

4 quarts " 1 gallon, gal. 

31 J gallons " 1 barrel, bbl. 

2 barrels, or 63 gal. " 1 hogshead,.. hhd. 

UNIT EQUIVALENTS. 

pt. gi. 
qt. 1=4 

gal . 1 = 2 = 8 

bbl . 1 = 4 = 8 = 32 

hnd . 1 = 31J = 126 = 252 = 1008 

1 = 2 = 63 = 252 = 504 = 2016 

Scale — ascending, 4, 2, 4, 31 J, 2; descending, 2, 31 J, 4, 2, 4. 

The following denominations are also in use : 

42 gallons make 1 tierce. 

2 hogsheads, or 120 gallons, " 1 pipe or butt. 
2 pipes or 4 hogsheads, " 1 tun. 

Notes. — 1. The denominations, barrel and hogshead, are used in estimating 
the capacity of cisterns, reservoirs, vats, etc. In Massachusetts the barrel is 
estimated at 32 gallons. 

2. The tierce, hogshead, pipe, butt, and tun are the names of casks, and do 
not express any fixed or definite measures. They are usually gauged, and have 
their capacities in gallons marked on them. Several of these denominations are 
still in use in England, (327—330). 



MEASURES OF CAPACITY. 1fl\ 

BEER MEASURE. 

291, Beer Measure is a species of liquid measure used in 
measuring beer, ale, and milk. 
The unit is the gallon. 

TABLE. 

2 pints (pt.) make 1 quart, qt. 

4 quarts " 1 gallon, gal. 

36 gallons " 1 barrel, bbl. 

1} barrels, or 54 gallons, " 1 hogshead, ..hhd. 

UNIT EQUIVALENTS. 

qt. pt. 

gal. 1 = 2 

bbl. 1=4=8 

hhd . 1 = 36 = 144 = 288 

1 = 1J = 54 = 216 = 432 

Scale — ascending, 2, 4, 36, 1}; descending, 1J, 36, 4, 2. 

This measure is not a standard j it is rapidly falling into disuse. 

II. Dry Measure. 

292* Dry Measure is used in measuring articles not liquid ; 

as grain, fruit, salt, roots, ashes, etc. 

The unit is the bushel, of which all the other denominations in 

the table are divisors. 

table. 

2 pints (pt.) make 1 quart, qt. 

8 quarts " 1 peck, pk. 

4 pecks - " 1 bushel, .. bu. or bush. 

UNIT EQUIVALENTS. 

qt. pt. 

Pk. 1=2 

bu. 1 = 8 = 16 

1 = 4 = 32 = 64 

Scale — ascending, 2, 8, 4 ; descending, 4, 8, 2. 

WEIGHTS. 
293. Weight is the measure of the quantity of matter a body 
contains, determined by the force of gravity. 

Note. — The process by which the quantity of matter or the force of gravity 
is obtained is called Weighing ; and consists in comparing the thing to Ioq 
weighed with some conventional standard. 



172 COMPOUND NUMBERS. 

Three scales of weight are used in the United States ; namely, 
Troy, Avoirdupois, and Apothecaries'. 

I. Troy Weight. 

394. Troy Weight is used in weighing gold, silver, and 
jewels; in philosophical experiments, and generally where great 
accuracy is required. 

The unit is the pound, and of this all the other denominations 
in the table are divisors. 

TABLE. 

24 grains (gr.) make 1 pennyweight, .. pwt. or dwt. 

20 pennyweights " 1 ounce, oz. 

12 ounces " 1 pound, lb. 

UNIT EQUIVALENTS. 

pwt. gr. 

oz. 1 = 24 

lb. 1 = 20 = 480 

1 = 12 -= 240 = 5760 

Scale — ascending, 24, 20, 12; descending, 12, 20, 24. 

Note. — Troy weight is sometimes called Goldsmith's Weight. 

II. Avoirdupois Weight. 

295. Avoirdupois Weight is used for all the ordinary pur- 
poses of weighing. 

The unit is the pound, and the table is made up of its divisors 
and multiples. 

TABLE. 

16 drams (dr.) make 1 ounce, oz. 

16 ounces " 1 pound, lb. 

100 lb. " 1 hundred weight, .. cwt. 
20 cwt., or 2000 lbs., " 1 ton, T. 

UNIT EQUIVALENTS. 

oz. dr. 

lb. 1 = 16 

cwt. 1 = 16 = 256 

T . 1 = 100 = 1600 = 25600 

1 = 20 = 2000 = 32000 = 512000 

Scale— -ascending, 16, 16, 100, 20; descending, 20, 100, 16, 16. 



WEIGHTS. 



173 



Note. — The long or gross ton, hundred weight, and quarter were formerly in 
common use; but they are now seldom used except in estimating English goods 
at the U. S. custom-houses, in freighting and wholesaling coal from the Penn- 
sylvania mines, and in the wholesale iron and plaster trade. 

LONG TON TABLE. 

28 lb. make 1 quarter, marked qr. 

4 qr. = 112 lb. " 1 hundred weight, " cwt. 
20 cwt. = 2240 lb. " 1 ton, " T. 

Scale — ascending, 28, 4, 20; descending, 20, 4, 28. 

296. The weight of the bushel of certain grains and roots 
has been fixed by statute in many of the States ; and these statute 
weights must govern in buying and selling, unless specific agree- 
ments to the contrary be made. 

TABLE OF AVOIRDUPOIS POUNDS IN A BUSHEL, 

As prescribed by statute in the several States named. 



COMMODITIES. 



Barley. . 



Blue Grass Seed 

Buckwheat 

Castor Beans 

Clover Seed 

Dried Apples 

Dried Peaches... 

Flax Seed.... 

Hair 

Hemp Seed 

Indian Corn 

Ind. Corn in ear 
Ind. Corn Meal. 
Mineral Coal 1 ... 

Oats 

Onions 

Peas 

Potatoes 

Kye 

Bye Meal 

Salt 2 

Timothy Seed- 
Wheat 

Wheat Bran 



5 

'5 

H 

o 


"5 

a 

a 
a 
O 


i 

a 


'3 
a 


s 
.5 


o 


O 

3 

a 


a 




£ 


1 


a 

i 


O 

a 
1 


o 

'I 


6 

a. 
S 

X 


4) 


o 


O 


c 
o 

6 


.2 

'5 

> 

a? 

c 
s 


•a" 
s 

o 

« 


o 

s 


h 

B 
O 


50 






48 
60 


48 
60 


48 
60 


48 
60 


32 




46 


48 


48 


48 
60 




48 


48 
62 


48 


46 


47 




46 


45 








14 


14 


14 


14 










14 




















40 


45 




40 
46 


50 
46 


52 

46 


52 






46 


42 


42 


52 

46 




50 


48 




42 


48 




46 


42 








60 


60 


60 


60 








60 


60 


60 




64 


60 


60 


60 








60 








24 


25 


24 










28 


28 


24 










28 








28 








33 


33 


33 










28 


28 


33 










28 








28 








56 


56 


56 


56 












56 




55 


55 


56 


















8 










ii 


































44 


44 


44 


44 












44 




















52 


56 


56 


52 
70 

48 
80 


56 
68 
50 
70 


56 

68 


56 
50 


56 


50 


56 
50 


56 


56 


52 

80 




56 


58 


56 


56 


56 


50 


56 


56 


32 


28 




32 

57 


32 

48 


35 
57 


33^ 

57 


32 


30 


30 
52 


32 


32 


35 

57 


30 


30 


32 

60 


32 


34 


32 


50 


32 


36 
50 




60 




60 


60 


60 


60 




60 








60 


60 


60 


60 




60 




60 


60 


60 


54 


56 




54 
45 


56 

50 
45 


56 

50 
45 


56 

50 
45 


32 


50 


56 

50 


56 

i 


56 


56 

50 
45 




56 


56 

56 
44 


56 


56 


56 


50 


56 


56 


60 


56 


60 


60 
20 


60 


60 
20 


60 
20 


60 




60 


60 
[ 


60 


60 
20 




60 


60 


60 


60 


60 




60 


60 



.48 



! 42 



; 28 
56 



i 56 



i 32 



i 56 



46 

i 60 



1 In Kentucky, 80 lbs. of bituminous coal or 70 lbs. of cannel coal make 1 bushel. 

3 In Pennsylvania, 80 lbs. coarse, 70 lbs. ground, or 62 lbs. fine salt make 1 bushel ; and 
in Illinois, 50 lbs. common or 55 lbs. fine salt make 1 bushel. 
, * In Maine, 64 lbs. of ruta baga turnips or beets make 1 bushel. 

15* 



174 



COMPOUND NUMBERS. 



Notes. ~ 1. The weight of a barrel of flour is 7 quarters of old, or long ton 
weight. 

2. The weight of a bushel of Indian corn and rye, as adopted by most of the 
States, and of a bushel of salt is 2 quarters ; and of a barrel of salt 10 quarters, 
or £ of a long ton. 

The following denominations are also in use : 

56 pounds make 1 firkin of butter. 

quintal of dried salt fish, 
cask of raisins, 
barrel of flour. 

" " beef, pork, or fish. 

" " salt at the N. Y* State salt works. 



100 " 


" 1 


100 


" 1 


196 


" 1 


200 " 


" 1 


280 "■ 


" 1 



III. Apothecaries' Weight. 

297. Apothecaries' Weight is used by apothecaries and phy- 
sicians in compounding medicines ; but medicines are bought and 
sold by avoirdupois weight. 

The unit is the pound, of which all the other denominations in 
the table are divisors. 

TABLE. 

20 grains (gr.) make 1 scruple, sc. or 9. 

3 scruples " 1 dram, dr. or g. 

8 drams " 1 ounce, . . oz. or §. 

12 ounces " 1 pound, lb. or ft). 



lb. 

1 



UNIT EQUIVALENTS. 

sc. gr. 

dr. 1 = 20 

oz . 1 = 3 = 60 

1 = 8 = 24 = 480 

12 = 96 = 288 = 5760 



Scale — ascending, 20, 3, 8, 12 ; descending, 12, 8, 3, 20. 

apothecaries' fluid measure. 

SJ98t The measures for fluids, as adopted by apothecaries and 
physicians in the United States, to be used in compounding medi- 
cines, and putting them up for market, are given in the following 

TABLE. 
60 minims, (^V) make 1 fluidrachm, f£. 



8 fluidrachms, 
16 fluidounces, 
8 pints, 



1 fluidounce, f^. 

1 pint, O. 

1 gallon, Cong. 



TIME. 



175 



UNIT EQUIVALENTS. 

ft I = 60 

o. 1 = 8 = 480 

Cong. 1 = 16 = 128 = 1024 
1 = 8 = 128 = 2048 = 16384 
Scale — ascending, 60, 8, 16, 8 ; descending, 8, 16, 8, 60. 

MEASURE OF TIME. 
299, Time is the measure of duration. The unit is the day, 
and the table is made up of its divisors and multiples. 

TABLE. 

60 seconds (sec.) make 1 minute, . .min. 

60 minutes, " 1 hour, -♦ . h. 

24 hours, " 1 day, ~ da. 

7 days, " 1 week, wk. 

365 days, " 1 common year, ....... yr. 

366 days, " 1 leap year,.. yr. 

12 calendar months, " 1 year, yr. 

100 years, " 1 century, C. 

UNIT EQUIVALENTS. 

min. sec. 

h. 1 = 60 

da. 1 == 60 = 3600 

wk. 1 = 24 = 1440 = 86400 

1 = 7 = 168 = 10080 = 604800 

yr mo. (365 = 8760 = 525600 = 31536000 

1 = 12 = {366 = 8784 = 527040 = 31622400 

Scale — ascending, 60, 60, 24, 7 ; descending, 7, 24, 60, 60. 

The calendar year is divided as follows : — 

No. of month. Season. Names of months. Abbreviations. No of days. 

1 WintPr {January, Jan. 31 

2 Wmter, j February> Fe b. 28 or 29 

3 [March, Mar. 31 

4 Spring, < April, Apr. 30 

5 I May, 31 

6 J June, Jun. 30 

7 Summer, -j July, 31 

8 I August, Aug. 31 

9 ( September, Sept. 30 

10 Autumn, < October, Oct. 31 

11 (November, Nov. 30 

12 Winter, December, Dec. 31 
Notes. — 1. In most business transactions 30 days are called 1 month. 

2. The civil day begins and ends at 12 o'clock; midnight. The aatronomi- 



17$ COMPOUND NUMBERS. 

cal day, used by astronomers in dating events, begins and ends at 12 o'clock, 
noon. The civil year is composed of civil days. 

BISSEXTILE OR LEAP YEAR. 

300* The period of time required by the sun to pass from 
one vernal equinox to another, called the vernal or tropical year, 
is exactly 365 da. 5 h. 48 min. 49.7 sec. This is the true year, 
and it exceeds the common year by 5 h. 48 min. 49.7 sec. 

If 365 days be reckoned as 1 year; the time lost in the calendar 
will be 

In 1 yr., 5 h. 48 min. 49.7 sec. 
" 4 " 23 " 15 " 18.8 " 

The time thus lost in 4 years will lack only 44 min. 41.2 sec. of 
1 entire day. Hence, 

If every fourth year be reckoned as leap year, the time gained in 
the calender will be, 

In 4 yr., 44 min. 41.2 sec. 

"100// ( = 25x41yr.) 18 h. 37 " 10 " 

The time thus gained in 100 years will lack only 5 h. 22 min. 50 
sec. of 1 day. Hence 

If every fourth year be reckoned as leap year, the centennial years 
excepted, the time lost in the calendar will be, 

In 100 yr., 5 h. 22 min. 50 sec. 
" 400 " 21 " 31 " 20 " 

The time thus lost in 400 years lacks only 2 h. 28 min. 40 sec. of 1 
day. Hence 

If every fourth year be reckoned as leap year, 3 of every 4 cen- 
tennial years excepted, the time gained in the calendar will be, 

In 400 yr., 2 h. 28 min. 40 sec. 
" 4000 " 24 h. 46 min. 40 sec. 

The following rule for leap year will therefore render the calendar 
correct to within 1 day, for a period of 4000 years. 

I. Every year that is exactly divisible by 4 is a leap year, 
the centennial years excepted ; the other years are common years. 

II. Every centennial year that is exactly divisible by 400 is a 
leap year ; the other centennial years are common years. 

Notes. — 1. Julius Caesar, the Roman Emperor, decreed that the year should 
consist of 365 days 6 hours ; that the 6 hours should be disregarded for 3 suc- 
cessive years, and an entire day be added to every fourth year. This day was 
inserted in the calendar between the 24th and 25th days of February, and is 
called the intercalary day. As the Romans counted the days backward from the 
first day of the following month, the 24th of February was called by them sexto 



CIRCULAR MEASURE. 177 

calendas Martii, the sixth before the calends of March. The intercalary day 
which followed this was called bissexto calendas Martii; hence the name 
bissextile. 

2. In 1582 the error in the calendar as established by Julius Caesar had in- 
creased to 10 days; that is, too much time had been reckoned as a year, until 
the civil year was 10 days behind the solar year. To correct this error, Pope 
Gregory decreed that 10 entire days should be stricken from the calendar, and 
that the day following the 3d day of October, 1582, should be the 14th. This 
brought the vernal equinox at March 21 — the date on which it occurred in the 
year 325, at the time of the Council of Nice. 

3. The year as established by Julius Caesar is sometimes called the Julian 
year; and the period of time in which it was in force, namely from 46 years 
B. C. to 1582, is called the Julian Period. 

4. The year as established by Pope Gregory is called the Gregorian year, and 
the calendar now used is the Gregorian Calendar. 

5. Most Catholic countries adopted the Gregorian Calendar soon after it was 
established. Great Britain, however, continued to use the Julian Calendar until 
1752. At this time the civil year was 11 days behind the solar year. To cor- 
rect this error, the British Government decreed that 11 days should be stricken 
from the calendar, and that the day following the 2d day of September, 1752, 
should be the 14th. 

6. Time before the adoption of the Gregorian Calendar is called Old Style 
(0. S), and since, New Style, (N. S.) In Old Style the year commenced March 
25, and in New Style it commences January 1. 

7. Russia still reckons time by Old Style, or the Julian Calendar; hence their 
dates are now 12 days behind ours. 

8. The centuries are numbered from the commencement of the Christian era ; 
the months from the commencement of the year; the days from the commence- 
ment of the month, and the hours from the commencement of the day, (12 
o'clock, midnight.) Thus, May 23, 1860, 9 o'clock A. M., is the 9th hour of the 
23d day of the 5th month of the 60th year of the 19th century. 

MEASURE OF ANGLES. 

301. Circular Measure, or Circular Motion, is used princi- 
pally in surveying, navigation, astronomy, and geography, for 
reckoning latitude and longitude, determining locations of places 
and vessels, and computing difference of time. 

Every circle, great or small, is divisible into the same number 
of equal parts : as quarters, called quadrants ; twelfths, called signs ; 
360ths, called degrees, etc. Consequently the parts of different 
circles, although having the same names, are of different lengths. 

The unit is the degree, which is 3^ part of the space about a 
point in any plane. The table is made up of divisors and multiples 
of this unit. 

TABLE. 

60 seconds (") make 1 minute, . . . . / . 

60 minutes " 1 degree, °. 

30 degrees " 1 sign, S. 

12 signs, or 360°, " 1 circle, C. 

M 



Ifg COMPOUND NUMBERS- 

unit EQUIVALENTS. 

/ // 

1 = 60 

s l = 60 = 3600 

c 1 = 30 = 1800 = 108000 

1 = 12 = 360 = 21600 = 1296000 

Scale — ascending, 60, 60, 30, 12 ; descending, 12, 30, 60, 60. 

Notes. — 1. Minutes of the earth's circumference are called geographic or 
nautical miles. 

2. The denomination, signs, is confined exclusively to Astronomy. 

3. A degree has no fixed linear extent. When applied to any circle it is always 
_|^ part of the circumference. But, strictly speaking, it is not any part of a 
circle. 

4. 90° make a quadrant or right-angle; 
60° " " sextant " & of a circle. 



MISCELLANEOUS TABLES. 
303. COUNTING. 

12 units or things make 1 dozen. 
12 dozen " 1 gross. 

12 gross " 1 great gross. 

20 units " 1 score. 

303. PAPER. 

24 sheets make 1 quire. 

20 quires " 1 ream. 

2 reams " 1 bundle, 

5 bundles " 1 bale. 

304. BOOKS. 

The terms folio, quarto, octavo, duodecimo, etc., indicate the 
number of leaves into which a sheet of paper is folded. 



A sheet folded in 2 leaves 


is called 


a folio. 


A sheet folded in 4 leaves 


a 


a quarto, or 4to. 


A sheet folded in 8 leaves 


a 


an octavo, or 8vo. 


A sheet folded in 12 leaves 


a 


a 12mo. 


A sheet folded in 16 leaves 


<« 


a 16mo. 


A sheet folded in 18 leaves 


<< 


an 18mo. 


A sheet folded in 24 leaves 


tt 


a 24mo. 


A sheet folded in 32 leaves 


it 


a 32mo. 


305. COPYING. 




72 words make 1 folio or 


sheet of 


common law. 


90 " " 1 " " 


it a 


chancery. 



GOVERNMENT STANDARDS. 179 

GOVERNMENT STANDARDS 
OF MEASURES AND WEIGHTS. 

300. In early times, almost every province and chief city had 
its own measures and weights ; but these were neither definite nor 
uniform. This variety in the weights and measures of different 
countries has always proved a serious embarrassment to commerce ) 
hence the many attempts that have been made in modern times to 
establish uniformity. 

The English, American, and French Governments, in establish- 
ing their standards of measures and weights, founded them upon 
unalterable principles or laws of nature, as will be seen by ex- 
amining the several standards. 

UNITED STATES STANDARDS. 

307. In the year 1834 the U. S. Government adopted a uni- 
form standard of weights and measures, for the use of the custom 
houses, and the other branches of business connected with the 
General Government. Most of the States which have adopted 
any standards have taken those of the General Government. 

308* The invariable standard unit from which the standard 
units of measure and weight are derived is the day. 

Astronomers have proved that the diurnal revolution of the 
earth is entirely uniform, always performing equal parts of a revo- 
lution on its axis in equal periods of duration. 

Having decided upon the invariable standard unit, a measure 
of this unit was sought that should in some manner be connected 
with extension as well as with this unit. A clock pendulum 
whose rod is of any given length, is found always to vibrate the 
same number of times in the same period of duration. Having 
now the day and the pendulum, the different standards hereafter 
given have been determined and adopted, 

STANDARD OF EXTENSION. 

300* The U. JS. standard unit of measures of extension, whether 
linear, superficial, or solid, is the yard of 3 feet, or 36 inches, 



180 COMPOUND NUMBERS. 

and is the same as the Imperial standard yard of Great Britain. 
It is determined as follows : The rod of a pendulum vibrating 
seconds of mean time, in the latitude of London, in a vacuum, at 
the level of the sea, is divided into 391393 equal parts, and 360000 
of these parts are 36 inches, or 1 standard yard. Hence, such a 
pendulum rod is 39.1393 inches long, and the standard yard is 
fff§§| of the length of the pendulum rod. 

STANDARDS OF CAPACITY. 

310* The U. S. standard unit of liquid measure is the old 
English wine gallon, of 231 cubic inches, which is equal to 
8.33888 pounds avoirdupois of distilled water at its maximum 
density; that is, at the temperature of 39.83° Fahrenheit, the ba- 
rometer at 30 inches. 

311* The TJ. $. standard unit of dry measure is the British 

Winchester bushel, which is 18J inches in diameter and 8 inches 

deep, and contains 2150.42 cubic inches, equal to 77.6274 pounds 

avoirdupois of distilled water, at its maximum density. A gallon, 

dry measure, contains 268.8 cubic inches. 

Notes. — 1. Grain and some other commodities are sold by stricken measure^ 
and in such cases the " measure is to be stricken with a round stick or roller, 
straight, and of the same diameter from end to end." 

2. Coal, ashes, marl, manure, corn in the ear, fruit and roots are sold by heap 
measure. The bushel, heap measure, is the Winchester bushel heaped in the 
form of a cone, which cone must be 19£ inches in diameter (= to the outside 
diameter of the standard bushel measure,) and at least 6 inches high. A bushel, 
heap measure, contains 2747.7167 cubic inches, or 597.2967 cubic inches more 
than a bushel stricken measure. Since 1 peck contains iiso.a = 537.605 cubic 
inches, the bushel, heap measure, contains 59.6917 cubic inches more than 5 
pecks. As this is about 1 bu. 1 pk. If pt., it is sufficiently accurate in practice, 
to call 5 pecks stricken measure a heap bushel. 

3. A standard bushel, stricken measure, is commonly estimated at 2150.4 
cubic inches. The old English standard bushel from which the United States 
standard bushel was derived, was kept at Winchester, England; hence the name. 

4. The wine and dry measures of the same denomination are of different capac- 
ities. The exact and the relative size of each may be readily seen by the fol- 
lowing 

312* COMPARATIVE TABLE OP MEASURES OF CAPACITY. 

Cubic in. in 
one gallon. 

Wine measure, 231 

Dry measure (i pk.,) . . 268£ 

Note. — The beer gallon of 282 inches is retained in use only by custom. 



Cubic in. in 


Cubic in. in 


Cubic in. in 


one quart. 


one pint. 


one gill. 


57| 


281 


7 7 

' S 2 


67i 


33§ 


8! 



GOVERNMENT STANDARDS. Igl 

STANDARDS OP WEIGHT. 

313. It has been found that a given volume or quantity of 
distilled rain water at a given temperature always weighs the same. 
Hence, a cubic inch of distilled rain water has been adopted as 
the standard of weight. 

tlH. The U. S. standard unit of weight is the Troy pound of 
the Mint, which is the same as the Imperial standard pound of 
Great Britain, and is determined as follows : A cubic inch of dis- 
tilled water in a vacuum, weighed by brass weights, also in a 
vacuum, at a temperature of 62° Fahrenheit's thermometer, is 
equal to 252.458 grains, of which the standard Troy pound con- 
tains 5760. 

315 • The U. S. Avoirdupois pound is determined from the 
standard Troy pound, and contains 7000 Troy grains. Hence, 
the Troy pound is f ££g = f|f °f an avoirdupois pound. But 
the Troy ounce contains 5 ^|° = 480 grains, and the avoirdupois 
ounce 7 £g° = 437.5 grains; and an ounce Troy is 480 — 437.5 
= 42.5 grains greater than an ounce avoirdupois. The pound, 
ounce, and grain, Apothecaries' weight, are the same as the like 
denominations in Troy weight, the only difference in the two 
tables being in the divisions of the ounce. 

316. COMPARATIVE TABLE OF WEIGHTS. 

Troy. Avoirdupois. Apothecaries'. 

1 pound = 5760 grains, = 7000 grains. = 5760 grains, 
1 ounce = 480 " = 437.5 " = 480 " 



48U M = 43/. O " = 48U 

175 pounds, = 144 pounds. = 175 pounds, 



STANDARD SETS OF WEIGHTS AND MEASURES. 

317* A uniform set of weights and measures for all the States 
was approved by Congress, June 14, 1836, and furnished to the 
States in 1842. The set furnished consisted of 

A yard. 

A set of Troy weights. 

A set of Avoirdupois weights. 



Jg2 COMPOUND NUMBERS. 

A wine gallon, and its subdivisions. 
A half bushel, and its subdivisions. 

318. State Sealers of Weights and Measures furnish standard 
sets of weights and measures to counties and towns. 
A County Standard consists of 

1. A large balance, comprising a brass beam and scale dishes, 
with stand and lever. 

2. A small balance, with a drawer stand for small weights. 

3. A set of large brass weights, namely, 50, 20, 10, and 5 lb. 

4. A set of small brass weights, avoirdupois, namely, 4, 2, and 
1 lb., 8, 4, 2, 1, i, and i oz. 

5. A brass yard measure, graduated to feet and inches, and the 
first foot graduated to eighths of an inch, and also decimally ; with 
a graduation to cloth measure on the opposite side ; in a case. 

6. A set of liquid measures, made of copper, namely, 1 gal., J 
gal., 1 qt., 1 pt., i pt., 1 gi.; in a case. 

7. A set of dry measures, of copper, namely, } bu., 1 pk.> \ pk. 
(or 1 gal.), 2 qt. (or J gal.), 1 qt.; in a case. 

ENGLISH MEASURES AND WEIGHTS. 
GOVERNMENT STANDARDS. 

319* The English act establishing standard measures and 
weights, called " The Act of Uniformity," took effect Jan. 1, 1826, 
and the standards then adopted, form what is called the Imperial 
System. 

320. The Invariable Standard Unit of this system is the 
same as that of the United States, and is described in the Act of 
Uniformity as follows : " Take a pendulum which will vibrate 
seconds in London, on a level of the sea, in a vacuum ; divide all 
that part thereof which lies between the axis of suspension and 
the center of oscillation, into 391393 equal parts ; then will 10000 
of those parts be an imperial inch, 12 whereof make a foot, and 
36 whereof make a yard." 



ENGLISH MEASURES AND WEIGHTS. 183 

STANDARD OF EXTENSION. 

33S1. The English Standard Unit of Measures of Extension, 
whether linear, superficial, or solid, is identical with that of the 
United States, (309). 

STANDARDS OF CAPACITY. 

322. The imperial Standard Gallon, for liquids and all dry 
substances, is a measure that will contain 10 pounds avoirdupois 
weight of distilled water, weighed in air, at 62° Fahrenheit, the 
barometer at 30 inches. It contains 277.274 cubic inches. 

323. The Imperial Standard Bushel is equal to 8 gallons or 
80 pounds of distilled water, weighed in the manner above de- 
scribed. It contains 2218.192 cubic inches. 

STANDARDS OF WEIGHT 

324:. The Imperial Standard Pound is the pound Troy, 
which is identical with that of the United States Standard Troy 
pound of the Mint, (314.) 

32d. The Imperial Avoirdupois Pound contains 7000 Troy 
grains, and the Troy pound 5760. It also is identical with the 
United States avoirdupois pound. 

TABLES. 

336. The denominations in the standard tables of measures 
of extension, capacity, and weights, are the same in Great Britain 
and the United States. But some denominations in several of the 
tables are in use in various parts of Great Britain that are not 
known in the United States. 

These denominations are retained in use by common consent, 
and are recognized by the English common law. They are as fol- 
lows : 

337. MEASURES OF EXTENSION. 



18 inches make 1 cubit. 

45 inches or 
5 quarters of the standard yard 



| " 1 ell. 



Note. — The cubit was originally the length of a man's forearm and hand; or 
the distance from the elbow to the end of the middle finger. 



Jg4 COMPOUND NUMBERS. 

328. MEASURES OP CAPACITY. 
LIQUID MEASURES. 

9 old ale gallons make 1 firkin. 

4 firkins " 1 barrel of beer. 



7 i Imperial " "1 firkin. 

52J Imperial gallons or * 
63 wine 

70 Imperial gallons or 
84 wine 

2 hogsheads, that is 
105 Imperial gallons or 
126 wine 
2 pipes " 1 tun 



1 hogshead. 

1 puncheon or 
i of a tun, 

1 pipe. 



Pipes of wine are of different capacities, as follows : 

110 wine gallons make 1 pipe of Madeira. 

( Barcelona, 
120 " " 1 " ^Vidonia, or 

( Teneriffe. 
130 " " 1 " Sherry. 

138 " " 1 " Port. 

14n u u -, « (Bucellas, or 

i4U X | Lisbon. 

339. DRY MEASURE. 

8 bushels of 70 pounds each make 1 quarter of wheat. 
36 " heaped measure, " 1 chaldron of coal. 

Note. — The quarter of wheat is 560 pounds, or I of a ton of 2240 pounds. 

330. WEIGHTS. 

8 pounds of butchers' meat make 1 stone. 

14 " " other commodities " 1 " or } of a cwt. 

2 stone, or 28 pounds " 1 todd of wool. 

70 pounds of salt " 1 bushel. 

Note. — The English quarter is 28 pounds, the hundred weight is 112 pounds, 
and the ton is 20 hundred weight, or 2240 pounds. 



FRENCH MEASURES AND WEIGHTS. 
GOVERNMENT STANDARDS. 

331. The tables of standard measures and weights adopted 
by the French Government are all formed upon a decimal scale, 
and constitute what is called the French Metrical System. 



FRENCH MEASURES AND WEIGHTS. 185 

332. Invariable Standard Unit. The French metrical sys- 
tem has, for its unit of all measures, whether of length, area, 
solidity, capacity, or weight, a uniform invariable standard, adopted 
from nature and called the mitre. It was determined and estab- 
lished as follows : a very accurate survey of that portion of the 
terrestrial meridian, or north and south circle, between Dunkirk 
and Barcelona, France, was made, under the direction of Govern- 
ment, and from this measurement the exact length of a quadrant 
of the entire meridian, or the distance from the equator to the 
north pole, was computed. The ten millionth part of this arc was 
denominated a metre, and from this all the standard units of 
measure and weight are derived and determined. 

STANDARDS OF EXTENSION. 

333. The French Standard Linear Unit is the metre. 
334* The French Standard Unit of Area is the Are, which 

is a unit 10 metres square, and contains 100 square metres. 

335* The French Standard Unit of Solidity and Capacity 
is the Litre, which is the cube of the tenth part of the metre. 

STANDARD OF WEIGHT. 

336. The French Standard Unit of Weight is the Gramme, 
which is determined as follows : the weight in a vacuum of a 
cubic decimetre or litre of distilled water, at its maximum density, 
was called a kilogramme, and the thousandth part of this was 
called a Gramme, and was declared to be the unit of weight. 

NOMENCLATURE OF THE TABLES. 

337* It has already been remarked, (330), that the tables are 
all formed upon a decimal scale. The names of the multiples and 
divisors of the Government standard units in the tables are formed, 
by combining the names of the standard units with prefixes ; the 
names of the multiples being formed by employing the prefixes 
deca, (ten), hecto, (hundred), kilo, (thousand), and myria, (ten 
thousand), taken from the Greek numerals ; and the names of the 
divisors by employing the prefixes deci, (tenth), centi, (hundredth), 
16* 



186 COMPOUND NUMBERS. 

mill, (thousandth), from the Latin numerals. Hence the name 
of any denomination indicates whether a unit of that denomination 
is greater or less than the standard unit of the table. 

338. I. French Linear Measure. 



J 

10 millimetres 


make 1 centimetre. 


10 centimetres 


it 


1 decimetre. 


10 decimetres 


a 


1 metre. 


10 metres 


a 


1 decametre. 


10 decametres 


u 


1 hectometre. 


10 hectometres 


it 


1 kilometre. 


10 kilometres 


a 


1 myriametre 



Notes. — 1. The metre is equal to 39.3685 inches, the standard rod of brass 
on which the former is measured being at the temperature of 32° Fahrenheit, 
and the English standard brass yard or " Scale of Troughton" at 62°. Hence, a 
metre is equal to 3.2807 feet English measure. 

2. The length of a metre being 39.3685 inches, and of a clock pendulum 
vibrating seconds at the level of the sea in the latitude of London 39.1393 
inches, the two standards differ only .2292, or less than £ of an inch. 

339. II. French Square Measure. 

table. 

100 square metres, or centiares (10 metres square) make 1 are. 
100 ares (10 ares square) " 1 hectoare. 

Note. — A square metre or centiare is equal to 1.19589444 square yards, and 
an are to 119.589444 square yards. 

340. III. French Liquid and Dry Measure. 

table. 

10 decilitres make 1 litre. 

10 litres • " 1 decalitre. 

10 decalitres " 1 hectolitre. 

10 hectolitres " 1 kilolitre. 

Notes. — 1. A litre is equal to 61.53294 cubic inches, or 1.06552 quarts of aU. 
S. liquid gallon. 

2. A table of Solid or Cubic Measure is also in use in some parts of France, 
although it is not established or regulated by government enactments or decrees. 
The unit of this table is a cubic metre, which is equal to 61532.94238 cubic 
inches, or 35.60934 cubic feet. This unit is called a Stere. 

TABLE. 

10 decisteres make 1 stere. 
10 steres " 1 decastere. 



MONEY AND CURRENCIES. 187 

341. IV. French Weight. 

TABLE. 

10 milligrammes make 1 centigramme. 

10 centigrammes " 1 decigramme. 

10 decigrammes " 1 gramme. 

10 grammes " 1 decagramme. 

10 decagrammes " 1 hectogramme. 

10 hectogrammes " 1 kilogramme. 

100 kilogrammes " 1 quintal. 

10 quintals " JlmiUier.or 

^ (1 ton of sea water. 

Notes. — 1. A gramme is equal to 15.433159 Troy grains. 
2. A kilogramme is equal to 2 lb. 8 oz. 3 pwt. 1.159 gr. Troy, or 2 lb. 3 oz. 
4.1549 dr. Avoirdupois. 

342. Comparative Table of the United States, English, 
and French Standard Units of Measures and Weights. 

United States. English. French. 

Extension, Yd. of 3 ft., or 36 in. Same as U. S. Metre, 39.3685 in. 

rano .. ) Wine gal., 231 cu. in. Imp'l gal., 277.274 cu. in. Litre, 61.53294 cu. in. 

capacity, | Whlch > r bu>? 2150.42 eu. in. Imp'l bu., 2218.192 cu. in. 

Weight, Troy lb., 5760 gr. Imperial lb., 5760 gr. Gramme, 15.433159 T. gr. 

Notes. — 1. An Imperial^gallon is equal to 1.2 wine gallons. 

2. An old ale or beer gallon is very nearly 1.221 wine gallons, or 1.017 Im- 
perial gallons. 

3. In ordinary computations 2150.4 cu. in. may be taken as » Winchester 
bushel, and 2218.2 cu. in. as an Imperial bushel. 



MONEY AND CURRENCIES. 

343. Money is the commodity adopted to serve as the uni- 
versal equivalent or measure of value of all other commodities, 
and for which individuals readily exchange their surplus products 
or their services. 

344. Coin is metal struck, stamped, or pressed with a die, to 

give it a legal, fixed value, for the purpose of circulating as 

money. 

Note. — Tbe coins of civilized nations consist of gold, silver, copper, and 
nickel. 

345. A Mint is a place in which the coin of a country or 
government is manufactured. 

Note. — In all civilized countries mints and coinage are under tbe exclusive 
direction and control of government. 



188 



COMPOUND NUMBERS. 



340. An Alloy is a compound of any metal with another of 

less value. In coinage, the less valuable or baser metal is not 

reckoned of any value. 

Note. — Gold and silver, in their pure state, are too soft and flexible for coin- 
age; hence they are hardened by compounding them with an alloy of baser 
metal, while their color and other valuable qualities are not materially impaired. 

347. An Assayer is a person who determines the composi- 
tion and consequent value of alloys of gold and silver. 

The fineness of gold is estimated by carats, as follows : — 

Any mass or quantity of gold, either pure or alloyed, is divided 
into 24 equal parts, and each part is called a carat. 

Fine gold is pure, and is 24 carats fine. 

Alloyed gold is as many carats fine as it contains parts in 24 of 
fine or pure gold. Thus, gold 20 carats fine contains 20 parts or 
carats of fine gold, and 4 parts or carats of alloy. 

348. An Ingot is a small mass or bar of gold or silver, in- 
tended either for coinage or exportation. Ingots for exportation 
usually have the assayer's or mint value stamped upon them. 

340. Bullion is uncoined gold or silver. 

3t>0. Bank Bills or Bank Notes are bills or notes issued by 
a banking company, and are payable to the bearer in gold or silver, 
at the bank, on demand. They are substitutes for coin, but are 
not legal tender in payment of debts or other obligations. 

331. Treasury Notes are notes issued by the General Govern- 
ment, and are payable to the bearer in gold or silver, at the gene- 
ral treasury, at a specified time. 

352. Currency is coin, bank bills, treasury notes, and other 
substitutes for money, employed in trade and commerce. 

353. A Circulating Medium is the currency or money of a 
country or government. 

354. A Decimal Currency is a currency whose denomna- 
tions increase and decrease according to the decimal scale. 

I. United States Money. 

355. The currency of the United States is decimal currency, 
and is sometimes called Federal Money. 



MONEY AND CURRENCIES. 



189 



The unit is the dollar, and all the other denominations are either 
^ivisors or multiples of this unit. 

TABLE. 

10 mills (m.) make 1 cent, .ct. 

10 cents " 1 dime, d. 

10 dimes " 1 dollar, $. 

10 dollars " 1 eagle, E. 

UNIT EQUIVALENTS. 

,ct. m. 

d. 1 = 10 

$ 1 = 10 = 100 

B . 1 = 10 == 100 = 1000 

1 = 10 = 100 = 1000 = 10000 

Scale — . uniformly 10. 

Notes. — 1. Federal Money was adopted by Congress in 1786. 
2. The character $ is supposed to be a contraction of U. S., (United States,) 
the U being placed upon the S. 

Coins. The gold coins are the double eagle, eagle, half eagle, 
quarter eagle, three dollar piece and dollar. 

The silver coins are the half and quarter dollar, dime and half 
dime, and three-cent piece. 

The nickel coin is the cent. 

Notes. — 1. The following pieces of gold are in circulation, but are not legal 
coin, viz. : the fifty dollar piece, and the half and quarter dollar pieces. 

2. The silver dollar, and the copper cent and half cent, are no longer coined 
for general circulation. 

3. The mill is a denomination used only in computations; it is not a coin. 

3o0. Government Standard. By Act of Congress, January 
18, 1837, all gold and silver coins must consist of 9 parts (.900) 
pure metal, and 1 part (.100) alloy. The alloy for gold must con- 
sist of equal parts of silver and copper, and the alloy for silver of 
pure copper. 

The three-cent piece is 3 parts (f ) silver, and 1 part (J) copper. 

The nickel cent is 88 parts copper and 12 parts nickel. 

STATE CURRENCIES. 

So 7« United States money is reckoned in dollars, dimes, cents, 
and mills, one dollar being uniformly valued in all the States at 
100 cents ; but in many of the States money is sometimes reckoned 
in dollars, shillings, and pence. 



J90 COMPOUND NUMBERS. 

Note. — At the time of the adoption of our decimal currency by Congress, in 
1786, the colonial currency, or bills of credit, issued by the colonies, had depre- 
ciated in value, and this depreciation, being unequal in the different colonies, 
gave rise to the different values of the State currencies; this variation continues 
wherever the denominations of shillings and pence are in use. 

Georgia Currency. 
Georgia, South Carolina, $1 = 4s. 8d. = 56d. 

Canada Currency. 

Canada, Nova Scotia, $1 = 5s. = 60d. 

New England Currency. 

New England States, Indiana, Illinois, 1 

Missouri, Virginia, Kentucky, Tennes- V $1 = 6s. = 72d. 

see, Mississippi, Texas, J 

Pennsylvania Currency. 

New Jersey, Pennsylvania, Delaware, ) $1 = 7 S 6d = 90d 

Maryland, j • • • • $ 

New York Currency. 

New York, Ohio, Michigan, ) .$1 = 8s. = 96d. 

North Carolina, j 



II. Canada Money. 
338. The currency of the Canadian provinces is decimal, and 
the table and denominations are the same as those of the United 
States money. 

Note. — The decimal currency was adopted by the Canadian Parliament in 
1858, and the Act took effect in 1859. Previous to the latter year the money 
of Canada was reckoned in pounds, shillings, and pence, the same as in Eng- 
land. 

Coins. The new Canadian coins are of silver and copper. 

The silver coins are the shilling or 20-cent piece, the dime, and 

half dime. 

The copper coin is the cent. 

Note. — The 20-cent piece represents the value of the shilling of the old 
Canada Currency. 

359. Government Standard. The silver coins consist of 92 

parts (.925) pure silver and 75 parts (.075) copper. That is, they 

are .925 fine. 

Note. — The value of the 20-cent piece in United States money is 18§ cents, 
of the dime 9J cents, and of the half dime 4§ cents. 



MONEY AND CURRENCIES. \Q\ 

III. English Money. 

360. English or Sterling Money is the currency of Great 
Britain. 

The unit is the pound sterling, and all the other denominations 
are divisors of this unit. 

TABLE. 

4 farthings (far. or qr.) make 1 penny, d. 

12 pence " 1 shilling, . . . s. 

20 shillings " 1 pound or sovereign .» £ or sov. 

UNIT EQUIVALENTS. 

d. far. 

8. 1= 4 

X, or sov. 1 = 12 = 48 

1 = 20 = 240 = 960 
Scale — ascending, 4, 12, 20 ; descending, 20, 12, 4. 

Notes. — 1. Farthings are generally expressed as fractions of a penny; thus, 

1 far., sometimes called 1 quarter, (qr.) = id. ; 3 far. =|d. 

2. The old/, the original abbreviation for shillings, was formerly written be- 
tween shillings and pence, and d, the abbreviation for pence, was omitted. Thus 
2s. 6d. was written 2/6. A straight line is now used in place of the/, and shil- 
lings are written on the right of it and pence on the left. Thus, 2/6, 10/3, 
etc. 

Coins. The gold coins are the sovereign (= £1) and the half 
sovereign, (= 10s.) 

The silver coins are the crown (= 5s.), the half crown (= 2s. 
6d.), the shilling, and the 6 penny piece. 

The copper coins are the penny, half penny, and farthing. 

Note. — The guinea (= 21s.) and the half guinea (= 10s. 6d. sterling) are old 
gold coins, that are still in circulation, but are no longer coined. 

361. Government Standard. The standard fineness of Eng- 
lish gold coin is 11 parts pure gold and 1 part alloy ; that is, it is 
22 carats fine. The standard fineness of silver coin is 11 oz. 

2 pwt. (= 11.1 oz.) pure silver to 18 pwt. (= .9 oz.) alloy. Hence 
the silver coins are 11 oz. 2 pwt. fine ; that is, 11 oz. 2 pwt. pure 
silver in 1 lb. standard silver. 

This standard is 87 parts (f £ = .925) pure silver and 3 parts 

(4 3 o = - 075 ) C0 PP er - 

Note. — A pound of English standard gold is equal in value to 14.2878 lb. = 
14 lb. 3 oz. 9 pwt. 1.727 gr. of silver. 



192 COMPOUND NUMBERS. 

IV. French Money. 

362. The currency of France is decimal currency. 
The unit is the franc, of which the other denominations are 
divisors. 

TABLE. 

10 millimes make 1 centime. 
10 centimes " 1 franc. 
Scale — uniformly 10. 

Coins. The gold coin is the 20-franc piece, or Louis. 
The silver coins are the franc and the demi franc. 

Note. — In France accounts are kept in francs and decimes. A franc is equal 
to 18.6 cents U. S. money. 



363. COMPARATIVE TABLE OF MONEYS. 



English, 
lqr. 
Id. 
Is. 

4s. Id. 2 T Vbqr. 
£1 



U.S. 


French. 


= $ .oo^ij 

= '02^ 
= .242 
= 1.00 

c= 4.84 


1 millime = 
1 centime = 
1 franc = 



U.S. 

$ .00186 
.0186 
.186 



REDUCTION. 

364. Reduction is the process of changing a number from 
one denomination to another without altering its value. 

Reduction is of two kinds, Descending and Ascending. 

365. Reduction Descending is changing a number of one 
denomination to another denomination of less unit value; thus, 
$1 = 10 dimes = 100 cents = 1000 mills. 

366. Reduction Ascending is changing a number of one 
denomination to another denomination of greater unit value; thus, 
1000 mills == 100 eents = 10 dimes = $1. 

REDUCTION DESCENDING. 
CASE I. 

367. To reduce a compound number to lower de- 
nominations. 

1. Reduce 3 mi. 1 fur. 17 rd. 2 yd. 1 ft. 8. in. to inches. 






REDUCTION. 



193 



OPERATION. 



3 mi. 1 fur. 17 rd. 2 yd. 1 ft. 8 in. 



25 
40 



fur. 



1017 rd. 
_5* 

5087 
_508J 

5595|yd. 
3 



16787J ft. 
12 

201458 in. 



Analysis. Since in 
1 mi. there are 8 fur., 
in 3 miles there are 
3X8 fur. = 24 fur., 
and the 1 fur. in the 
given number, added, 
makes 25 fur. in 3 mi. 
1 fur. Since in 1 fur. 
there are 40 rd., in 25 
fur. there are 25 X 40 
rd. == 1000 rd., and 
the 17 rd. in the given 
number added, makes 
1017 rd. in 3 mi. 1 fur. 
17 rd. Since in 1 rd. 
there are 5 J yd., in 
1017 rd. there are 
1017 X 5 J yd. = 5593 J yd., which plus the 2 yd. in the given number 
= 5595J yd. in 3 mi. 1 fur. 17 rd. 2 yd. Since in 1 yd. there are 
3 ft., in 5595J yd. there are 5595 J X 3 ft. = 16786J ft., which plus 
the 1 ft. in the given number = 16787} ft. in 3 mi. 1 fur. 17 rd. 2 yd. 
1 ft. And since in 1 ft. there are 12 in., in 16787 J ft. there are 
16787} X 12 in. = 201450 in., which plus the 8 in. in the given num- 
ber = 201458 in. in the given compound number. On examining the 
operation, we find that we have successively multiplied the numbers 
in the descending scale of linear measure from miles to inches, inclu- 
sive. But, as either factor may be used as a multiplicand, (82, 1), we 
may consider the numbers in the descending scale as multipliers. 
Hence the following 

Rule. I. Multiply the highest denomination of the given 
corn-pound number by that number of the scale which will reduce it 
to the next lower denomination, and add to the product the given 
number, if any, of that lower denomination. 

II. Proceed, in the same manner with the results obtained in each 
lower denomination, until the reduction is brought to the denomina- 
tion required. 

EXAMPLES FOR PRACTICE. 

1. In 16 lb. 10 oz. 18 pwt. 5 gr., how many grains? 
17 



194 COMPOUND NUMBERS. 

2. In £133 6 s. 9d., how many farthings? Ans. 128,000. 

3. Change 100 mi. to inches. Ans. 6336000 in. 

4. How many rods of fence will inclose a farm 1} miles 
square ? Ans. 1920 rd. 

5. The grey limestone of Central New York weighs 175 lbs. 
to the cubic foot; what is the weight of a block 8 ft. long and 
1 yd. square ? Ans. 6 T. 6 cwt. 

6. What will be the cost of 1 hhd. of molasses at $.28 per gal. ? 

7. A man wishes to ship 1548 bu. 1 pk. of potatoes in barrels 
containing 2 bu. 3 pk. each ; how many barrels must he obtain ? 

8. A grocer bought] 10 bu. of chestnuts at $3.75 a bushel, and 
retailed them at $.06i a pint; how much was his whole gain ? 

9. Reduce 90° 17' 40" to seconds. Ans. 325060". 

10. In the 18th century how many days ? Ans. 36524 da. 

11. At 6J cts. each ; what will be the cost of a great-gross of 
writing books ? Ans. $108. 

12. How large an edition of an octavo book can be printed 
from 4 bales 4 bundles 1 ream 10 quires of paper, allowing 8 
sheets to the volume ? Ans. 2970 vol. 

13. Suppose your age to be 18 yr. 24 da. ; how many minutes 
old are you, allowing 4 leap years to have occurred in that time ? 

14. How many pence in 481 sovereigns ? Ans. 115,440 d. 

15. Reduce $7f to mills. Ans. 7375 mills. 

16. In 3 P. of Sherry wine, how many qt. ? Ans. 1560 qt. 

17. Reduce 37 Eng. ells 1 qr. to yd. Ans. 46 yd. 2 qr. 

18. In £6 10s. lOd. how many dollars U. S. currency ? 

19. Reduce 6;0. 14f£ 3fj 45 n to minims. 

20. Reduce 1 T. 1 P. 1 hhd. to Imperial gallons. 

Ans. 367} Imperial gal. 

21. How many dollars Canada currency are equal to £126 12s. 
6d.? Ans. $506}. 

22. How many bottles containing pints, quarts, and two quarts, 
of each an equal number, may be filled from a hogshead of wine ? 

Ans. 72. 

23. How many steps of 2 ft. 9 in. each, will a man take, in 
walking from Erie to Cleveland, the distance being 95 mi. ? 



REDUCTION. 195 

24. A grocer bought 12 bbl. of cider at $lf a barrel, and after 
converting it into vinegar, he retailed it at 6 cents a quart ; how 
much was his whole gain ? Ans. $69.72. 

25. In 75 A. 4 sq. ch. 18 P. 118 sq. 1. how man y square links? 

26. How many inches high is a horse that measures 16 hands ? 

27. If a vessel sail 150 leagues in a day, how many statute 
miles does she sail ? Ans. 517.5. 

28. If 14 A. be sold from a field containing 50 A., how many 
square rods will the remainder contain ? Ans. 5,760 sq. rd. 

29. A man returning from Pike's Peak has 86 lb. 8 oz. of 
pure gold ; what is its value at $1.04^ per pwt. ? Ans. 9169.60. 

30. A person having 8 hhd. of tobacco, each weighing 9 cwt. 
42 lb., wishes to put it into boxes containing 48 lb. each ; how 
many boxes must he obtain ? Ans. 157. 

31. A merchant bought 12 bbl. of salt at $1J a barrel, and re- 
tailed it at f of a cent a pound ; how much was his whole gain ? 

32. A physician bought IK) 10^ of quinine at $2.25 an ounce, 
and dealt it out in doses of 10 gr. at $.12 i each ; how much more 
than cost did he receive? Ans. $82.50. 



CASE II. 

368. To reduce a denominate fraction from a greater 
to a less unit. 

1. Reduce fa of a gallon to the fraction of a gill. 

operation. Analysis. To re- 

fa gal. X i X f X | = T 8 T gi. duce g allons t0 giUs, 

Or, 



11, 



we multiply succes- 
sively by 4, 2, and 4, 

1 the numbers in the de- 
4 scending scale. And 

2 since the given nuni- 

w ber is a fraction, we 

8 = -&- gi., Ans. indicate the process, 

as in multiplication of 
fractions, after which we perform the indicated operations, and ob- 
tain T 8 T , the answer. Hence, 



11 



196 COMPOUND NUMBERS. 

Rule. Multiply the fraction of the higher denomination by the 
numbers in the descending scale successively, between the given and 
the required denomination. 
Note. — Cancellation may be applied wherever practicable. 

EXAMPLES FOR PRACTICE. 

1. Reduce 3^ of a lb. Troy to the fraction of a pennyweight. 

Ans. I pwt. 

2. Reduce g ^ of a hhd. to the fraction of a pint. 

3. Reduce ^yy^ of a mile to the fraction of a yard. 

Ans. I yd. 

4. Reduce gf 3 of a gallon to the fraction of a gill. 

5. What part of a dram is 5 qV o" of f of | of T 6 T of 8^ pounds 
avoirdupois weight ? Ans. ^\^% dr. 

6. Reduce t |q of a cent to the fraction of a dollar. 

7. Reduce ^ of a rod to the fraction of a link. Ans. | 1. 

8. Reduce ^ of a grain to the fraction of a scruple. 

9. What fraction of a yard is | of T 4 T of a rod ? 

10. T 5 g of -a week is § of how many days? Ans. 8| da. 

11. What fraction of a square rod is y-g 3 ^ of 4| times T 2 ^ of an 
acre ? Ans. T 3 ^ sq. rd. 

CASE III. 

369. To reduce a denominate fraction to integers 

of lower denominations. 

1. What is the value of | of a bushel ? 

operation. Analysis, f bu. = § of 

I bu. X 4 = f pk. = If pk. 4 pk., or 1? pk.; § pk. = § 

3 p k. x 8 = % 4 qt. = 4| qt. of 8 qt. == 4| qt. ; and f qt. 

fqt. X2= f pt. = lfpt. =jof 2pt. = 15pt. The 

1 pk. 4 qt. 1| pt., Ans. units > l P k " 4 «*•' X P*" 

with the last denominate 

fraction, § pt., form the answer. Hence, 

Rule. I. Multiply the fraction by that number in the scale 

which will reduce it to the next lower denomination, and if the 

result be an improper fraction, reduce it to a whole or mixed 

number. 



REDUCTION. 197 

II. Proceed with the fractional part, if any, as before, until 
reduced to the denominations required. 

III. The units of the several denominations , arranged in their 
order, will be the required result. 

EXAMPLES FOR PRACTICE. 

1. Reduce T 9 ^ of a yard to integers of lower denominations. 

Ans. 2 ft. 8| in. 

2. Reduce § of a month to lower denominations. 

3. Reduce ||^ of a short ton to lower denominations. 

4. What is the value of § of a long ton ? 

Ans. 11 cwt. 12 lb. 7| oz. 

5. What is the value of f of 2| pounds apothecaries' weight ? 

6. What is the value of T \ of an acre ? 

Ans. 2 R. 6 P. 4 sq. yd. 5 sq. ft. 127 T 5 3 sq. in. 

7. Reduce | of a mile to integers of lower denominations. 

8. What is the value of | of a great gross ? 

Ans. 6 gross 10 doz. 3|. 

9. What is the value in geographic miles of T 9 ^ of a great 
circle ? Ans. 12150 mi. 

10. What is the value of § of 3f cords of wood ? 

Ans. 2 Cd. 5 cd. ft. 9f cu. ft. 

11. The distance from Buffalo to Cincinnati is 438 miles; hav- 
ing traveled § of this distance, how far have I yet to travel ? 

Ans. 262 mi. 6 fur. 16 rd. 

12. What is the value of || fg ? Ans. 3 f ^ 35 ttj,. 

13. What is the value of | of a sign ? 

Ans. 12° 51' 25"f . 

14. A man having a hogshead of wine, sold 6 T 6 3 of it; how 
much remained ? Ans. 33 gal. 3 qt. 1 pt. 1 T \ gi. 



CASE IV. 

370. To reduce a denominate decimal to integers 
of lower denominations. 

1. Reduce .125 of a barrel to integers of lower denominations. 
17* 



198 COMPOUND NUMBERS. 










OPERATION. 








.125 

31.5 








3.9375 gal. 
4 








3.7500 qt. 
2 

1.50 pt. 
4 

2.0 gi. 




3 


gal. 


3 qt. 1 pt. 2 gi., 


Ans. 



Analysis. We first multiply 
the given decimal, .125 of a barrel, 
by 31.5 (= 31J) to reduce it to 
gallons, and obtain 3.9375 gallons. 
Omitting the 3 gallons, we mul- 
tiply the decimal, .9375 gal., by 
4 to reduce it to quarts, and obtain 
3.75 quarts. We next multiply 
the decimal part of this result by 
2, to reduce it to pints, and obtain 
1.5 pints. And the decimal part 
of this result we multiply by 4 to 
reduce it to gills, and obtain 2 

gills. The integers of the several denominations, arranged in their 

order, form the answer. Hence, 

Rule. I. Multiply the given denominate decimal by that num- 
ber in the descending scale which will reduce it to the next lower 
denomination, and point off the result as in multiplication of 
decimals. 

II. Proceed with the decimal part of the product in the same 
manner until reduced to the required denominations. The integers 
at the left will be the answer required. 

EXAMPLES FOR PRACTICE. 

1. What is the value of .645 of a day ? 

Ans. 15 h. 28 min. 48 sec. 

2. What is the value of .765 of a pound Troy ? 

3. What is the value of .6625 of a mile? 

4. What is the value of .8469 of a degree ? 

Ans. 50' 48.84". 

5. What is the value of .875 of a hhd. ? 

6. What is the value of £.85251 ? 

Ans. 17 s. 2.4 -f far. 

7. What is the value of .715° ? Ans. 42' 54". 

8. What is the value of 7.88125 acres ? 

Arts. 7 A. 3 R. 21 P. 

9. What is the value of .625 of a fathom? Ans. 3| ft. 



REDUCTION. 



199 



10. What is the value of .375625 of a barrel of pork? 

11. What is the value of .1150390625 Cong. ? 

Ans. 14 f§ 5f^48 m. 

12. What is the value of .61 of a tun of wine ? 

Ans. 1 P. 13 gal. 3 qt. 1.76 gi. 

REDUCTION ASCENDING. 
CASE I. 

371. To reduce a denominate number to a com- 
pound number of higher denominations. 
1. Reduce 157540 minutes to weeks. 



OPERATION. 

60)157540 min. 

24 ) 2625 h. + 40 min. 

7)109 da. + 9h. 

15 wk. + 4 da. 
15 wk. 4 da. 9 h. 40 min., Ans. 



Analysis. Dividing 
the given number of 
minutes by 60, because 
there are ^ as many- 
hours as minutes, and we 
obtain 2645 h. plus a re- 
mainder of 40 min. We 
next divide the 2645 h. 
by 24, because there are 
2*! as many days as hours, and we find that 2645 h. = 109 da. plus a 
remainder of 9 h. Lastly we divide the 109 da. by 7, because there 
are \ as many weeks as days, and we find that 109 da. = 15 wk. plus 
a remainder of 4 da. The last quotient with the several remainders 
annexed in the order of the succeeding denominations, form the 
answer. 

2. Reduce 201458 inches to miles. 

OPERATION. 

12) 201458 in. 



3 ) 16788 ft. 2 in. 
5| or 5.5) 5596 yd. 

40) 1017 rd. 2 yd. lft. 6 in. 
8) 25 fur. 17 rd. 
3 mi. 1 fur. 
3 mi. 1 fur. 17 rd. 2 yd. 1 ft. 8 in., Ans. 



Analysis. We 
divide successively 
by the numbers in 
the ascending scale 
of linear measure, 
in the same manner 
as in the last pre- 
ceding operation. 
But, in dividing the 
5596 yd. by 5} or 
5.5, we have a re- 



200 COMPOUND NUMBERS. 

mainder of 2 J yd., and this reduced to its equivalent compound num- 
ber, (169) = 2 yd. 1 ft. 6 in. In forming our final result, the 6 in. 
of this number are added to the first remainder, 2 in., making the 8 in. 
as given in the answer. From these examples and analyses we deduce 
the following 

Rule. I. Divide the given concrete or denominate number by 
that number of the ascending scale which will reduce it to the next 
higher denomination. 

II. Divide the quotient by the next higher number in the scale ; 

and so proceed to the highest denomination required. The last 

quotient, with the several remainders annexed in a reversed order, 

will be the answer. 

Note. — The several corresponding cases in reduction descending and reduc- 
tion ascending, being opposites, mutually prove each other. 



EXAMPLES FOR PRACTICE. 

1. Reduce 1913551 drams to tons. 

2. In 97920 gr. of medicine how many lb. ? Ans. 17 lb. 






3. Reduce 1000000 in. to mi. 

4. How many acres in a field 120 rd. long and 56 rd. wide ? 

5. In a pile of wood 60 ft. long, 15 ft. wide, and 10 ft. high, 
how many cords ? Ans. 70 Cd. 2 cd. ft. 8 cu. ft. 

6. How many fathoms deep is a pond that measures 28 ft. 6 
in. ? Ans. 4£ fath. 

7. In 30876 gi. how many hhd. ? 

8. How many bushels of corn in 27072 qt. ? Ans. 846 bu. 

9. At 2 cts. a gill, how much alcohol may be bought for $2.54 ? 

10. In 1234567 far. how many £ ? Ans. £128 If d. 

11. Reduce 2468 pence to half crowns. 

12. In $88.35 how many francs ? Ans. 475. 

13. In 622080 cu. in. how many tons of round timber ? 

14. In 84621 \ how many Cong. ? 

15. If 135 million Gillott steel pens are manufactured yearly, 
how many great-gross will they make ? Ans. 78125. 

16. Reduce 1020300" to S. Ans. 9 S. 13° 25' 

17. In 411405 sec. how many da. ? 



REDUCTION. 201 

18. During a storm at sea, a ship changed her latitude 412 
geographic miles ; how many degrees and minutes did she change ? 

Ans. 6° 52'. 

19. If a man travel at the rate of a minute of distance in 20 
minutes of time, how much time would he require to travel round 
the earth ? Ans. 300 days. 

20. In 120 gross how many score ? Ans. 864. 

21. How many miles in the semi-circumference of the earth? 

22. How much time will a person gain in 36 yr. by rising 45 
min. earlier, and retiring 25 min. later, every day, allowing for 9 
leap years ? Ans. 182 da. 15 h. 

23. A grocer bought 20 gal. of milk by beer measure, and sold 
it by wine measure ; how many quarts did he gain ? Ans. 17f ^. 

24. How many bushels of oats in Connecticut are equivalent to 
1500 bushels in Iowa ? Ans. 1875 bu. 

25. Reduce 120 leagues to statute miles. Ans. 414 mi. 

26. In 1 bbl. 1 gal. 2 qt. wine measure, how many beer gal- 
lons? Ans. 27-g\. 

27. Reduce 150 U. S. bushels to Imperial bushels. 

Ans. 145.415 + Imp'l. bu. 

28. How many squares in a floor 68 ft. 8 in. long, and 33 ft. 
wide? Ans. 22§^. 

29. How many cubic inches in a solid 4 ft. long 3 ft. wide, and 
1 ft. 6 in. thick ? 

30. How many acres in a field 120 rd. long and 56 rd. wide ? 

31. Change 356 dr. apothecaries weight, to Troy weight. 

32. A coal dealer bought 175 tons of coal at S3. 75 by the long 
ton, and sold it at $4.50 by the short ton ; how much was his 
whole gain ? Ans. $210. 

33. How many acres of land can be purchased in the city of 
New York for $73750, at $1.25 a square foot? 

Ans. 1 A. 115 P. 1SH sq. ft. 

34. An Ohio farmer sold a load of corn weighing 2492 lb., and a 
load of wheat weighing 2175 lb. ; for the corn he received $.60 a 
bushel, and for the wheat $1.20 a bushel; how much did he re- 
ceive for both loads ? Ans. $70.20. 



202 



COMPOUND NUMBERS. 



The following examples are given to illustrate a short and prac- 
tical method of reducing currencies. 

35. What will be the cost of 54 bu. of corn at 5s. a bushel, 
New England currency ? 



OPERATION. 



Or, 



54 x 5 = 270s. 
270s. -r- 6 = $45 



5_ 

$45 



Analysis. Since 1 bu. costs 
5s., 54 bu. cost 54 X 5s. == 270s.; 
and since 6s. make $1 N. E. 
currency, 270 -~ 6 = $45, Ans. 



OPERATION. 



6 



100 



Or, % 



$300 



25 
2 

$300 



36. What will 270 bu. of wheat cost, @ 8s. 4d. Penn. currency ? 

Analysis. Multiply the 
quantity by the price in Penn. 
currency, and divide the cost 
by the value of $1 in the same 
currency; or reduce the shil- 
lings and pence to a fraction 
of a shilling, before multiply- 
ing and dividing. 

37. Bought 5 hhd. of rum at the rate of 2s. 4d. a quart, Geor- 
gia currency ; how much was the whole cost ? 

OPERATION. 

5 
63 

2 
4 

t 



5 

63 



Or, 



$630 



$630 



Analysis. In this ex- 
ample we first reduce 5 
hhd. to quarts by multiply- 
ing by 63 and 4, and then 
proceed as in the preceding 
examples. 



38. Sold 120 barrels of apples, each containing 2 bu. 2 pk., at 
4s. 7d. a bushel, and received pay in cloth at 10s. 5d. a yard ; how 
many yards of cloth, did I receive ? 

Analysis. The operation in this example is 
similar to the preceding examples, except that we 
divide the cost of the apples by the price of a unit 
of the article received in payment, reduced to units 
of the same denomination as the price of a unit of 
the article sold. The result will be the same in 
$132 whatever currency. 



OPERATION. 

12 

m 

11 



* 



REDUCTION. 203 

39. What cost 75 yards of flannel at 3s. 6d. per yard, New 
England currency 1 Ans. $43.75. 

40. A man in Philadelphia worked 5 weeks at 6s. 4d. a day ; 
how much did his wages amount to ? Ans. $25.33 J. 

41. A farmer exchanged 2 bushels of beans worth 10s. 6d. per 
bushel, for two kinds of sugar, the one at lOd. and the other at 
lid. per pound, taking the same quantity of each kind; how 
many pounds of sugar did he receive ? Ans. 24 lb. 

42. If corn be rated at 5s. lOd. per bushel in Vermont, at what 
price in the currency of New Jersey must it Jbe sold, in order to 
gain $7.50 on 54 bushels? 

CASE II. 

372. To reduce a denominate fraction from a less to 

a greater unit. 

1. Reduce T 8 T of a gill to the fraction of a gallon. 

operation. ^ Analysis. To re- 

t 8 t gi- X \ X £ X i = £ gal. duce S ills t0 S* 110118 ' 

we divide successively 



Or, 

11 
4 



44 



by 4, 2, and 4, the 
ff numbers in the as- 

cending scale. And 
since the given num- 

ber is a fraction, we 

1 = ^j, Ans. indicate the process, 

as in division of frac- 
tions, after which we perform the indicated operations, and obtain 
^k, the answer. Hence, 

Rule, Divide the fraction of the lower denomination by the 
numbers in the ascending scale successively , between the given and 
the required denomination. 

Note. — The operation may frequently fee shortened by cancellation. 

EXAMPLES FOR PRACTICE. 



1. Reduce § of a shilling to the fraction of a pound. 



Ans. £?£$. 



204 COMPOUND NUMBERS. 

2. Reduce f of a pennyweight to the fraction of a pound 
Troy. Ans. 3^ lb. 

3. What part of a ton is f of a pound avoirdupois weight ? 

4. What fraction of an hour is | of 20 seconds ? 

5. What is the fractional difference between 5^ of a hhd. 
and I of a pt. ? Ans. 3^^ hhd. 

6. 2 | 4 of I of § of a pint is what fraction of 2 pecks ? 

Ans. § 

7. Reduce f of | of T 9 ? of a cord foot to the fraction of a 
cord. Ans. ^ 3 H cd. 

8. What part of an acre is T 3 ^ of T \ of 9^ square rods ? 

9. I of 5^ furlongs is | of y 1 ^ of how many miles? 

Ans. 12f mi. 

10. A block of granite containing | of f of 20 1 cubic feet, is 
what fraction of a perch ? Ans. ±\ P. 

11. What part of a cord of wood is a pile 7<| ft. long, 2 ft. high, 
and 3| feet wide ? Ans. Iff Cd. 

12. Reduce f of an inch to the fraction of an Ell English. 



CASE III. 

373. To reduce a compound number to a fraction of 
a higher denomination. 

1. Reduce 2 oz. 12 pwt. 12 gr. to the fraction of a pound Troy. 

operation. Analysis. To find 

2 oz. 12 pwt. 12 gr. = 1260 gr. what part one compound 

1 lb. Troy = 5760 gr. number is of another, 

if iu lb - = 32 lb -> Ans - they must be like num ^ 

bers and reduced to the 

same denomination. In 2 oz. 12 pwt. 12 gr. there are 1260 gr., and 

in 1 lb. there are 5760 gr. Therefore 1 gr. is ^V^ lb., and 1260 gr. 

are iHi lb. = 52 lb., the answer. Hence, 

Rule. Reduce the given number to its lowest denomination for 

the numerator, and a unit of the required denomination to the same 

denomination for the denominator of the required fraction. 

Note. — If the given number contain a fraction, the denominator of this frac- 
tion must be regarded as the lowest denomination. 



REDUCTION. 205 

EXAMPLES FOR PRACTICE. 

1. Eeduce 2 R. 20 P. to the fraction of an acre. 

Ans. | A. 

2. What part of a mile is 6 fur. 26 rd. 3 yd. 2 ft. ? 

3. What part of a £ is 18s. 5d. 2 T % far. ? Ans. £i|. 

4. What part of 21 lb. Apothecaries' weight is 7J 7£ 29 14 
gr.? Ans. y 3 ^ lb. 

5. What part of 3 weeks is 4 da. 16 h. 30 min. ? 

6. Reduce If pecks to the fraction of a bushel. 

7. From a hogshead of molasses 28 gal. 2 qt. were drawn ; 
what part of the whole remained in the cask? Ans. ||. 

8. Reduce 4 bundles 6 quires 16 sheets of paper to the frac- 
tion of a bale. Ans. | of a bale. 

9. What part of 54 cords of wood is 4800 cubic feet ? 

10. What is the value of ^f| of a dollar? Ans. $6.30. 

11. Reduce 30. 3f § If z 36 ty to the fraction of a Cong. 

12. What part of a ton of shipping is 36 cu. ft. 864 cu. in. ? 

CASE IV. 

374. To reduce a compound number to a decimal of 
a higher denomination. 

1. Reduce 3 cd. ft. 8 cu. ft. to the decimal of a cord. 

operation. Analysis. We re- 

16 8.0 cu. ft. duce the 8 cu. ft. to 



3.5000 cd. ft. 



the decimal of a cd. 

ft., by annexing ci- 

.4375 Cd., Ans. phers, and dividing 

Or, by 16, the number of 

3 cd. ft. 8 cu. ft. = 56 cu. ft. cu. ft. in 1 cd. ft,, an- 

1 Cd. = 128 cu. ft. nexing the decimal 

J^ Cd. = T 7 g Cd. = .4375 Cd., Ans. quotient to the 3 cd. 

ft. We now reduce 
the 3. 5 cd. ft. to Cd. or a decimal of a Cd., by dividing by 8, the 
number of cd. ft. in 1 Cd., and we have .4375 Cd., the answer. 
Or, we may reduce the 3d cd. ft. 8 cu. ft., to the fraction of a Cd., 
18 



206 COMPOUND NUMBERS. 

(as in 373), and we shall have T \% Cd. = T \ Cd., which, reduced to 
its equivalent decimal, equals .4375 Cd., the same as before. Hence, 

Rule. Divide the lowest denomination given by that number 
in the scale which will reduce it to the next higher denomination, 
and annex the quotient as a decimal to that higher. Proceed in 
the same manner until the whole is reduced to the denomination 
required. Or, 

Seduce the given number to a fraction of the required denomi- 
nation, and reduce this fraction to a decimal. 

EXAMPLES FOR PRACTICE. 

1. Reduce 5 da. 9 h. 46 min. 48 sec. to the decimal of a week. 

Ans. .7725 wk. 

2. Reduce 3° 27' 46.44" to the decimal of a sign. 

3. Reduce 1 R. 11.52 P. to the decimal of an acre. 

4. What part of 4 oz. is 2 oz. 16 pwt. 19.2 gr ? Ans. .71. 
6. What part of a furlong is 28 rd. 2 yd. 1 ft. 11.04 in. ? 

6. Reduce 3|§ to the decimal of a pound. 

7. Reduce 126 A. 4 sq. ch. 12 P. to the decimal of a town- 
ship. Ans. .0054893 + £. 

8. What part of a fathom is 3f ft. ? Ans. .625 fath. 

9. What part of 1| bushels is .45 of a peck ? Ans. .09. 

10. What part of 3 A. 2 R. is 1 R. 11.52 P. ? Ans. .092. 

11. Reduce § of £ of 22f lb. to the decimal of a short ton. 

12. What part of a fg is 5 % 36 Tit ? Ans - 3 ?Z* 

13. Reduce 50 gal. 3 qt. 1 pt. to the decimal of a tun. 

Ans. .20188 + T. 
ADDITION. 

375. Compound numbers are added, subtracted, multiplied, 
and divided by the same general methods as are employed in 
simple numbers. The corresponding processes are based upon the 
same principles ; and the only modification of the operations and 
rules is that required for borrowing, carrying, and reducing by a 
varying, instead of a uniform scale. 

376. 1. What is the sum of 50 hhd. 32 gal. 3 qt. 1 pt., 2 
hhd. 19 gal. 1 pt., 15 hhd. 46£ gal., and 9 hhd. 39 gal. 2£ qt. ? 



ADDITION. 207 

operation. Analysis. Writing the numbers so that 

hhd. gal. qt. pt. units of the same denomination shall stand 

50 32 3 1 in the same column, we add the numbers 

2 19 1 of the right hand or lowest denomination, 

15 46 and fi n d the amount to be 3 pints, which is 

9 39 2 1 equal to 1 qt. 1 pt. We write the 1 pt. under 

78 11 3 1 tne column of pints, and add the 1 qt. to 

the column of quarts. The amount of the 

numbers of the next higher denomination is 7 qt., which is equal to 

1 gal. 3 qt. We write the 3 qt. under the column of quarts, and add 

the 1 gal. to the column of gallons. Adding the gallons, we find the 

amount to be 137 gal., equal to 2 hhd. 11 gal. Writing the 11 gal. 

under the gallons in the given numbers, we add the 2 hhd. to the 

column of hogsheads. Adding the hogsheads, we find the amount to 

be 78 hhd., which we write under the left hand denomination, as in 

simple numbers. 

2. What is the sum of T 7 ^ wk., | da., and | h. 

operation. Analysis. We 

T 7 ^ wk. = 4 da. 21 h. 36 min. first find the value 

£ da = 14 u 24 min. °^ eac ^ fr ac ti° n * n 

I h.' = 22" '30 sec. ^teger^ of less de- 

-^ nominations, (369), 

5 12 22 30 and then add the 

Or, resulting or equiva- 

| da. X 4 = A w k; lent compound num- 



i "• -* 24 



x| = A wl ; bers - 



/ n wk. + 3% wk. + ^ wk. = \\ wk; Or, we may re- 

\\ wk. = 5 da. 12 h. 22 min. 30 sec. d . uce the S iv(m frac " 

tions to fractions of 

the same denomination, (368 or 372), then add them, and find the 

value of their sum in lower denominations. 

377. From these examples and illustrations we derive the 
following 

Rule. I. If any of the numbers are denominate fractions, or 
if any of the denominations are mixed numbers, reduce the frac- 
tions to integers of lower denominations. 

II. Write the numbers so that those of the same unit value wiU 
stand in the same column. 

III. Beginning at the right hand, add each denomination as in 



208 COMPOUND NUMBERS. 

simple numbers, carrying to each succeeding denomination one for 

as many units as it takes of the denomination added } to make one 

of the next higher denomination. 

Note. — The pupil cannot fail to see that the principles involved in adding 
compound numbers are the same as those in addition of simple numbers ; and 
that the only difference consists in the different carrying units. 

EXAMPLES EQR PRACTICE. 

(2-) 

B>. 5 3 3 gr. 

10 8 5 1 8 

7 7 6 2 13 

5 11 7 

21 10 16 

12 12 2 3 

7 1 19 








(1-) 




lb. 


oz. 


pwt. 


gr. 


14 


6 


12 


13 


17 


5 


3 


12 


15 




9 


16 


2 


7 


15 


20 


13 


2 


1 


19 


4 


1 


5 


21 


66 


7 


19 


5 




(3.) 




fur. 


rd. 


ft. 


in. 


7 


26 


11 


9 


4 


16 


7 


11 




36 


14 


3 


1 


9 


2 


8 


5 




10 


1 


6 


2 


5 




1 


15 


13 


10 



58 


4 5 
(4-) 


2 


19 


A. 


B. P. 


sq.yd 


eij.ft 


140 


3 17 


27 


6 


320 


1 30 


14 


2 


111 


7 


3 




214 


2 15 


22 


7 


100 


3 


6 


1 


25 


1 36 




8 


104 


2 9 


1 


4 



5. Add 1 T. 17 cwt. 8 lb., 5 cwt. 29 lb. 8 oz., 1 cwt. 42 lb. 
6 oz., and 17 lb. 8 oz. Ans. 2 T. 3 cwt. 97 lb. 6 oz. 

6. Add 6 yd. 2 ft., 3 yd. 1 ft. 8 in., 1 ft. 10J in., 2 yd. 2 ft. 
6i in., 2 ft. 7 in., and 2 yd. 5 in. Ans. 16 yd. 2 ft. 1 in. 

7. Add 4 Cd. 7 cd. ft., 2 Cd. 2 cd. ft. 12 cu. ft., 6 cd. ft. 15 
cu. ft., 5 Cd. 3 cd. ft. 8 cu. ft., and 2 Cd. 1 cu. ft. 

8. What is the sum of If hhd. 42 gal. 3 qt. 1\ pt., i gal. 
2 qt. | pt., and 1.75 pt. ? Ans. 2 hhd. 23 gal. 2 qt. 3 gi. 

9. What is the sum of 145J A., 7 A. 2 R. 29J P., 1A.3R. 
16.5 P., and | A. ? Ans. 156 A. 39J P. 

10. Required the sum of 31 bu. 2 pk., lOg bu., 5 bu. 6£ qt., 
14 bu. 2.75 pk., and § pk. Ans. 61 bu. 1 pk. If pt. 



SUBTRACTION. 209 

11. Required the value of 42 yr. 1\ mo. -f 10 yr. 3 wk. 5 da. 
+ 9f mo. + 1 wk. 16 h. 40 min. + | mo. + 3f da. 

Ans. 53 yr. 7 mo. 9 da. 23 h. 52 min. 

12. Add 3 S. 22° 50', 24° 36' 25.7", 17' 18.2", 1 S. 3° 12' 
15.5", 12° 36' 17.8", and 57.3". Ans. 6 S. 3° 33' 14.5". 

13. How many units in 1A gross 7| doz., 3 gross 1| doz., | of 
a great gross, 6| doz., and 4 doz. 7 units ? Ans. 2183. 

14. What is the sum of 240 A. 6 sq. ch., 212.1875 sq. ch., 
and 5 sq. ch. lOf P.? Ans. 262 A. 3 sq. ch. 13.8 P. 

15. Add 31 Pch. 18 cu. ft., 84.6 cu. ft., § Pch., and §° ct. ft. 

16. Add $3f , $25|, $12g, $2§, and $2.54f . Ans. $47.0725. 

17. What is the sum of 3 ft> 5 § 4 z 2 9 17 gr., 2 ft 5 3 12 
gr.,43 2^19 16 gr.? Ans. 5ft 10^4^29 5 gr. 

18. AN. Y. farmer received $.60 a bushel for 4 loads of corn ; 
the first contained 42.4 bu., the second 2866 lb., the third 36| 
bu., and the fourth 39 bu. 29 lb. How much did he receive for 
the whole ? Ans. $100.89. 

19. Bought three loads of hay at $8 per ton. The first weighed 
1.125 T., the second If T., and the third 2500 pounds ; how much 
did the whole cost? Ans. $30.20. 

20. A man in digging a cellar removed 140| cu. yd. of earth, 
in digging a cistern 24.875 cu. yd., and in digging a drain 46 cu. 
yd. 20| cu. ft. What was the amount of earth removed, and how 
much the cost at 18 cts. a cu. yd. ? 

Ans. 212.425 cu. yd. removed; $38.24 — cost. 



SUBTRACTION. 

378. 1. From 18 lb. 5 oz. 4 pwt. 14 gr. take 10 lb. 6 oz. 10 
pwt. 8 gr. 

Analysis. Writing the subtra- 
hend under the minuend, placing 
units of the same denomination under 
each other, we subtract 8 gr. from 
10 14 6 14 & r# an( ^ wr i* e the remainder, 6 

■ gr., underneath. Since we cannot 
18* o 





OPERATION. 




lb. 


oz. 


pwt. 


gr- 


18 


5 


4 


14 


10 


6 


10 


. 8 



bar. 


gal. 


qt 


12 


15 


3 


7 


18 


1 



210 COMPOUND NUMBERS. 

subtract 10 pwt. from 4 pwt., we add 1 oz. or 20 pwt. to the 4 pwt., 
subtract 10 pwt. from the sum, and write the remainder, 14 pwt., 
underneath. Having added 20 pwt. or 1 oz. to the 6 oz. in the sub- 
trahend, we find that we cannot subtract the sum, 7 oz., from the 5 
oz. in the minuend ; we therefore add 1 lb. or 12 oz. to the 5 oz., sub- 
tract 7 oz. from the sum, and write the remainder, 10 oz., underneath. 
Adding 12 oz. or 1 lb. to the 10 lb. in the subtrahend, we subtract 
the sum, 11 lb., from the 18 lb. in the minuend, as in simple numbers, 
and write the remainder, 7 lb., underneath. 

2. From 12 bar. 15 gal. 3 qt. take 7 bar. 18 gal. 1 qt. 

Analysis. Proceeding as in the last 

operation, we obtain a remainder of 4 bar. 

28} gal. 2 qt. But,' J gal. = 2 qt., which 

added to the 2 qt. in the remainder makes 

4 28? 2 i g a i^ an< i this added to the 28 gal. makes 

4 29 29 gal. ; and the answer is 4 bar. 29 gal. 

3. From f of a rod subtract J of a yard. 

Analysis. We first 

find the value of each 
fraction in integers of 
lower deneminations, 
(369), and then sub- 
tract the less value 
from the greater. Or, 
we may reduce the 
| rd. — 7j 3 2 rd. = || rd. ; given fractions to frac- 

| rd. = 3 yd. 1 ft. 14 in. tions of the same de- 

nomination, subtract 
the less value from the greater, and find the value of the remainder 
in integers of lower denominations. 

379* From these illustrations we deduce the following 
Rule. I. If any of the numbers are denominate fractions, or 
if any of the denominations are mixed numbers, reduce the frac- 
tions to integers of lower denominations, 

II. Write the subtrahend under the minuend, so that units of the 
same denomination shall stand under each other. 

III. Beginning at the right hand, subtract each denomination 
separately, as in simple numbers. 





|rd. 


OPERATION. 

= 4 yd. ft. 4£ 
= 2 « 3 


in. 
« 

. 3 

• 22 




yd. 

rd. 
rd. 


3 1 1£ 
Or, 

x A = fyd. X T \ = 

°2 
3 r A 2 7 r A . 

22 ra. — %% ra. , 
= 3 yd. 1 ft. li in. 


rd 



SUBTRACTION. 211 

IV. If the number of any denomination in the subtrahend ex- 
ceed that of the same denomination in the minuend, add to the 
number in the minuend as many units as make one of the next 
higher denomination, and then subtract ; in this case add 1 to the 
next higher denomination of the subtrahend before subtracting. 
Proceed in the same manner with each denomination. 

EXAMPLES FOR PRACTICE. 





(1-) 










<2-) 


mi. 


fur. 


rd. 


a 


in. 




A. R. P. 


From 175 


3 


27 


11 


4 




320 3 26.4 


Take 59 


6 


10 


12 


9 




150 2 31.86 


Kern. 115 


5 


16 


15 


1 




170 34.54 


(3.) 












(*■) 


hhd. gal. 


qt. 






yr. 


mo. 


wk. da. h. 


5 36 


3i 






45 


i 


3 o m 


2 45 


H 






10 


9 


1 22 6.8 



5. Subtract 15 rd. 10 ft. 3} in. from 26 rd. 11 ft. 3 in. 

Ans. 11 rd. llf in. 

6. From 1 T. 11 cwt. 30 lbs. 6 oz. take 18 cwt. 45 lb. 

7. Subtract .659 wk. from 2 wk. 3| da. 

Ans. 1 wk. 6 da. 5 h. 17 min. 16| sec. 

8. From \\\ hhd. take .90625 gal. Ans. 32 gal. 

9. From f of 3f A. take 3 R. 12.56 P. 

10. Subtract ? % lb. Troy, from 10 lb. 8 oz. 8 pwt. 

11. From a pile of wood containing 36 Cd. 4 cd. ft., there was 
sold 10 Cd. 6 cd. ft. 12 cu. ft. ; how much remained ? 

12. From 5J barrels take | of a hogshead. 

Ans. 4 bbl. 11 gal. 1 qt. 

13. Subtract f ^ of a day from f of a week. 

Ans. 4 da. 49 min. 30 sec. 

14. From f of a gross subtract f of a dozen. Ans. 6| doz. 

15. From § of a mile take \\ of a rod. 

16. Subtract 2 A. 3 R. 5.76 P. from 5 A. 1 R. 24.24 P. 

Ans. 2 A. 2 R. 18.48 P. 



212 COMPOUND NUMBERS. 

17. Subtract .0625 bu. from f pk. Ans. 4 qt. 

18. From the sum of § of 365| da. and f of 5| wk. take 49^ 
min. Ans. 33 wk. 1 da. 1 h. lOf min. 

19. From the sum of § of 3| mi. and 17$ rd., take 5| fur. 

20. From 15 bbl. 3.25 gal. take 14 bbl. 24 gal. 3.54 qt. 

21. A farmer in Ohio having 200 bu. of barley, sold 3 loads, 
the first weighing 1457 lb., the second 1578 lb., and the third 
1420 lb. ; how many bushels had he left? Ans. 107 bu. 9 lb. 

22. Of a farm containing 200 acres two lots were reserved, one 
containing 50 A. 136.4 P. and the other 48 A. 123.3 P. ; the re- 
mainder was sold at $35 per acre. How much did it bring ? 

Ans. $3513.19=fc. 

23. An excavation 17 ft. long, 12 ft. wide, and 7 ft. deep is to 
be made for a cellar; after 471 cu. yd. 16 cu. ft. 972 cu. in. of 
earth have been removed, how much more still remains to be 
taken out ? Ans. 5 cu. yd. 7 cu. ft. 756 cu. in. 

24. From the sum of f lb., 4| oz., and 31| pwt., take the differ- 
ence between f oz. and | pwt. Ans. 1 lb. 3 oz. 8 pwt. 21 gr. 

Ql 

25. From the sum of 5 T \ A., § of 6| A., jJL R., and T \ of 

2 T \ P., take 4 A. 25 P. 12 gq. yd. 

Ans. 5 A. 3 R. 5 P. 6 sq. yd. 

380. To find the difference in dates. 

1. How many years, months, days and hours from 3 o'clock p. 
m. of June 15, 1852, to 10 o'clock A. m. of Feb. 22, 1860? 

operation. Analysis. Since the later of two dates 

yr. mo. da. h. always expresses the greater period of 

1860 2 22 10 time, we write the later date for a minu- 

end and the earlier date for a subtrahend, 

7 8 6 19 placing the denominations in the order of 

the descending scale from left to right, 

(300, Note 8). We then subtract by the rule for subtraction of 

compound numbers. 

When the exact number of days is required for any period not 
exceeding one ordinary year, it may be readily found by the fol- 
lowing 



SUBTRACTION. 



213 



TABLE, 

Showing the number of days from any day of one month to the same 
day of any other month within one year. 



FROM ANT 
DAY OF 



January,... 
February... 

March 

April 

May 

June , 

July. 

August 

September., 

October 

November., 
December... 



TO THE SAME DAY OF THE NEXT 



Jan. Feb. Mar. Apr. May June July. Aug. Sept. Oct. Nov. Dec, 



365 

334 

306 

275 

245 

214 

184 

153 

122 

92 

61 

31 



31 
365 
337 
306 
276 
245 
215 
184 
153 
123 
92 
62 



59 
28 
365 
335 
304 
273 
243 
212 
181 
151 
120 
90 



90 
59 
31 
365 
335 
304 
274 
243 
212 
182 
151 
121 



120 

89 

61 

30 

365 

334 

304 

273 

242 

212 

181 

151 



151 

120 

92 

61 

31 

365 

335 

304 

273 

243 

212 

182 



181 

150 

122 

91 

61 

30 

365 

334 

303 

273 

242 

212 



212 

181 

153 

122 

92 

61 

31 

365 

334 

304 

273 

243 



243 

212 

184 

153 

123 

92 

62 

31 

365 

335 

304 

274 



273 

242 

214 

183 

153 

122 

92 

61 

30 

365 

334 

304 



304 

273 

245 

214 

184 

153 

123 

92 

61 

31 

365 

335 



334 

303 

275 

244 

214 

183 

153 

122 

91 

61 

30 

365 



If the days of the different months are not the same, the num- 
ber of days of difference should be added when the earlier day 
belongs to the month from which we reckon, and subtracted when 
it belongs to the month to which we find the time. If the 29th 
of February is to be included in the time computed, one day 
must be added to the result. 



EXAMPLES FOR PRACTICE. 

1. War between England and America was commenced April 
19, 1775, and peace was restored January 20, 1783 ; how long did 
the war continue ? Ans. 7 yr. 9 mo. 1 da. 

2. The Pilgrims landed at Plymouth Dec. 22, 1620, and Gen. 
Washington was born Feb. 22, 1732 ; what was the difference in 
time between these events ? 

3. The first settlement made in the U. S. was at Jamestown, Va., 
May 23, 1607; how many years from that time to July 4, 1860 ? 

4. How long has a note to run ; dated Jan. 30, 1859, and made 
payable June 3, 1861 ? Ans. 2 yr. 4 mo. 3 da. 



214 COMPOUND NUMBERS. 

5. How many years, months, and days from your birthday to 
this date ? 

6. What length of time elapsed from 16 minutes past 10 o'clock, 
A. M., July 4, 1855, to 22 minutes before 8 o'clock, p. m., Dec. 
12, 1860 ? Ans. 1988 da. 9 h. 22 min. 

7. What length of time will elapse from 40 minutes 25 seconds 
past 12 o'clock, noon, April 21, 1860, to 4 minutes 36 seconds 
before 5 o'clock, A. M., Jan. 1, 1862 ? 

8. How many days from the 4th September, to the 27th of 
May following ? Ans. 265 da. 



MULTIPLICATION. 

381. 1. Multiply 5 mi. 4 fur. 18 rd. 15 ft. by 6. 

operation. Analysis. Writing the multi- 

plier under the lowest denomi- 
5 mi. 4 fur. 18 rd. 15 ft. nation of the mu iti p ii ca nd, we 

multiply each denomination in 

33 2 33 7 J the multiplicand separately in 

order from lowest to highest, as 
in simple numbers, and carry from lower denominations to higher, 
according to the ascending scale of the multiplicand, as in addition 
of compound numbers. Hence, 

Rule. I. Write the multiplier under the lowest denomination 
of the multiplicand. 

II. Multiply as in simple numbers, and carry as in addition of 

compound numbers. 

Notes. — 1. When the multiplier is large, and is a composite number, we may 
shorten the work by multiplying by the component factors. 

2. The multiplier must be an abstract number. 

3. If any of the denominations are mixed numbers, they may either be re- 
duced to integers of lower denominations before multiplying, or they may be 
multiplied as directed in 193. 

4. The multiplication of a denominate fraction is the most readily performed 
by 193, after which the product may be reduced to integers of lower denomina- 
tions by 369. 

382. As the work of multiplying by large prime numbers is 
somewhat tedious, the following method may often be so modified 
and adapted as to greatly shorten the operation. 



MULTIPLICATION. 



215 



1. How many bushels of grain in 47 bags, each containing 2 bu 
1 pk. 4 qt. ? 



FIRST OPERATION. 

47 =(5x9) + 2 
2 bu. 1 pk. 4 qt. x 2 
5 



11 bu. 3 pk. 4 qt. in 5 bags. 



106 bu. 3 pk. 4 qt. in 45 bags. 
4 " 3 " "2 « 

111 bu. 2 pk. 4 qt. " 47 " 

SECOND OPERATION. 

47=(6x8)_l 
2 bu. 1 pk. 4 qt. x 1 
6 



14 bu. 1 pk. 

8 



in 6 bags. 



114 bu. in 48 bags. 

2 " lpk.4 qt. « 1 bag. 

Ill bu. 2 pk. 4 qt. " 47 



Analysis. Multiplying 
the contents of 1 bag by 5, 
and the resulting product by 
9, we have the contents of 
45 bags, which is the com- 
posite number next less than 
the given prime number, 47. 
Next, multiplying the con- 
tents of 1 bag by 2, we have 
the contents of 2 bags, which, 
added to the contents of 45 
bags, gives us the contents 
of 45 + 2 = 47 bags. 

Or, we may multiply the 
contents of 1 bag by the fac- 
tors of the composite num- 
ber next greater than the 
given prime number, 47, and 
from the last product sub- 
tract the multiplicand. 



EXAMPLES FOR PRACTICE 

00 

T. cwt. lb. oz. 

12 15 27 9 
8 



Prod. 102 



A. 

7 



mi. 

14 



(20 

fur. rd. ft. 

6 36 14 
9 



R. 

1 



20 8 
(3.) 

P. sq.yd. sq.ft. 
33 21 7 
6 



133 6 11 10J 



(4-) 

Cd. cd.ft. cu.ffc. 

10 7 13 
12 



Multiply 34 bu. 3 pk. 6 qt. 1 pt. by 14. 
Multiply 4 lb. 10 oz. 18.7 pwt. by 27. 

Am. 132 lb. 7 oz. 4.9. pwt. 
Multiply 9 I 3 z 2 9 13 gr. by 35. 



216 COMPOUND NUMBERS. 

8. Multiply 5 gal. 2 qt. 1 pt. 3.25 gi. by 96. 

9. Multiply 78 A. 3 R. 15 P. 15 sq. yd. by 15f . 

Ans. 1235 A. 1 R. 2 P. 23J sq. yd. 

10. What is 73 times 9 cu. yd. 10 cu. ft. 1424 cu. in.? 

11. Multiply 2 lb. 8 oz. 13 pwt. 18 gr. by 59. 

12. Multiply 4 yd. 1 ft. 4.7 in. by 125. 

13. If 1 qt. 2 gi. of wine fill 1 bottle, how much will be re- 
quired to fill a gross of bottles of the same capacity ? 

14. Multiply 7 0. 10 f g 4 f 3 25 ifl, by 24. 

Ans. 22 Cong. 7 0. 13 f I 1 f^. 

15. Multiply 3 hhd. 43 gal. 2.6 gi. by 17. 

16. Multiply 9 T. 13 cwt. 1 qr. 10.5 lb. by 1.7. 

Note. — When the multiplier contains a decimal, the multiplicand may be re- 
duced to the lowest denomination mentioned, or the lower denominations to a 
decimal of the higher, before multiplying. The result can be reduced to the 
compound number required. 

Ans. 16 T. 8 cwt. 2 qr. 20.35 lb. 

17. If a pipe discharge 2 hhd. 23 gal. 2 qt. 1 pt. of water in 1 
hour, how much will it discharge in 4.8 hours, if the water flow 
with the same velocity? Ans. 11 hhd. 25 gal. 1 pt. 2.4 gi. 

18. What will be the value of 1 dozen gold cups, each cup 
weighing 9 oz. 13 pwt. 8 gr., at $212.38 a pound ? 

19. What cost 5 casks of wine, each containing 27 gal. 3 qt. 1 
pt., at $1.37* a gallon? Ans. $191.64+. 

20. A farmer sold 5 loads of oats, averaging 37 bu. 3 pk. 5 qt. 
each, at $.65 per bushel; how much money did he receive for the 
grain? Ans. $123.20—. 

DIVISION. 

383. 1. Divide 37 A. 1 R. 16 P. by 8. 

operation. Analysis. Writing the divisor on 

the left of the dividend, we divide 

) the highest denomination, and obtain 

4 2 27 a quotient of 4 A. and a remainder 

of 5 A. Writing the quotient under 

the denomination divided, we reduce the remainder to roods, making 

20 R., which added to the 1 R. of the dividend, equals 21 R. Dividing 

this, we have a quotient of 2 R. and a remainder of 5 R. Writing 



DIVISION. 217 

the 2 R. under the denomination divided, we reduce the remainder to 
rods, making 200 P., which added to the 16 P. of the dividend, equals 
216 P. Dividing this, we have a quotient of 27 P. and no remainder. 
2. Divide 111 bu. 2 pk. 4 qt. by 47. 

OPERATION. 

47) 111 bu. 2 pk. 4 qt. ( 2 bu. 1 pk. 4 qt. Analysis The 

94 

— divisor being large, 

17 bu. rem. and a prime num- 

_4 ber, we divide by 

70 pk. in 17 bu. 2 pk. lon g division, set- 

47 tm g down the 

whole work of sub- 
traci 
ing. 



23 pk. rem. 
o r tractmg and reduc- 

188 qt. in 23 pk. 4 qt. 

188 



From these examples and illustrations we derive the following 
R t tle. I. Divide the highest denomination as in simple num- 
bers, and each succeeding denomination in the same manner, if 
there be no remainder. 

II. If there be a remainder after dividing any denomination, 
reduce it to the next lower denomination, adding in the given num- 
ber of that denomination, if any, and divide as before. 

III. The several partial quotients will be the quotient required. 

Notes. — 1. When the divisor is large and is a composite number, we may 
shorten the work by dividing by the factors. 

2. When the divisor and dividend are both compound numbers, they must 
both be reduced to the same denomination before dividing, and then the process 
is the same as in simple numbers. 

3. The division of a denominate fraction is most readily performed by 195, 
after which the quotient may be reduced to its equivalent compound number, 
by 369. 

EXAMPLES FOR PRACTICE. 

(1.) (2 : ) 

£ 8. d. far. lb. oz. pwt. gr. 

5)62 7 7 3 9)56 6 17 6 

Quotient, 12 9 6 8 61b. 3oz. 8pw.l4gr. 

19 



218 COMPOUND NUMBERS. 



(3.) 








m 




hhd. gal. 


CLt. 


P t. 


'T. 


cwt. qr. 


lb. 


12)9 28 


2 




19)373 


19 2 


4 






49 2 1 19 13 2 16 

5. Divide 358 A. 1 R. 17 P. 6 sq. yd. 2 sq. ft. by 7. 

-4»s. 51 A. 31 P. 8 sq. ft. 

6. Divide 192 bu. 3 pk. 1 qt. 1 pt. by 9. 

7. Divide 336 yd. 4 ft. 31 in. by 21. Ans. 16 yd. 2| in. 

8. Divide 77 sq. yd. 5 sq. ft. 82 sq. in. by 13. 

Ans. 5 sq. yd. 8 sq. ft. 106 sq. in. 

9. Divide 678 cu. yd. 1 cu. ft. 1038.05 en. in. by 67. 

10. Divide 1986 mi. 3 fur. 20 rd. 1 yd. by 108. 

11. Divide 12 sq. mi. 1 R. 30 P. by 22£. 

Ans. 341 A. 1 R. 16| P. 

Note 4. — Observe that 22% = \ 5 ; hence, multiply by 2, and divide the result 
by 45. 

12. Divide 365 da. 6 h. by 240. 

13. Divide 3794 cu. yd. 20 cu. ft. 709^ cu. in. by 33f 

14. Divide 121 lb. 3g 2j 19 4 gr. by 13f . 

15. Divide 28° 51' 27.765" by 2.754. Ans. 10° 28' 42}". 

16. Divide 202 yd. 1 ft. 6f in. by f. 

Ans. 337 yd. 1 ft. 7i in. 

17. Divide 1950 da. 15 h. 15f min. by 100. 

18. If a town 4 miles square be divided equally into 124 farms, 
how much will each farm contain ? Ans. 82 A. 2 R. 12|| P. 

19. A cellar 48 ft. 6 in. long, 24 ft. wide, and 6} ft. deep, was 
excavated by 6 men in 8 days ; how many cubic yards did each 
man excavate daily? Ans. 5 cu. yd. 22 cu. ft. 1080 cu. in. 

20. How many casks, each containing 2 bu. 3 pk. 6 qt., can be 
filled from 356 bu. 3 pk. 5 qt. of cherries? Ans. 121 J. 



LONGITUDE AND TIME. 

JI84L Since the earth performs one complete revolution on its 
axis in a day or 24 hours, the sun appears to pass from east to 
west round the earth, or through 360° of longitude once in every 

I 

i! 



LONGITUDE AND TIME. 219 

24 hours of time. Hence the relation of time to the real motion 
of the earth or the apparent motion of the sun, is as follows : 

Time. Longitude. 

24 h. = 360° 

1 h. or 60 mm. = %%°° = 15° = 900' 

1 min. or 60 sec. == ||° = % 0/ = 15' = 900" 

lsec. == iV = %°" = 15" 

Hence, 1 h. of time = 15° of longitude. 

1 min. " = 15' " " 

1 sec. u = 15" " " 

CASE I. 

385. To find the difference of longitude between twc 
places or meridians, when the difference of time is 
known. 

Analysis. A difference of 1 h. of time corresponds to a difference 
of 15° of longitude, of 1 min. of time to a difference of 15' min. of 
longitude, and of 1 sec. of time to a difference of 15" of longitude, 
(184). Hence, the 

Rule. Multiply the difference of time, expressed in hours, 
minutes j and seconds, by 15, according to the rule for multiplica- 
tion of compound numbers ; the product will be the difference of 
longitude in degrees, minutes, and seconds. 

: Notes. — 1. If one place be in east, and the other in west longitude, the dif- 
ference of longitude is found by adding them, and if the sum be greater than 
180°, it must be subtracted from 360°. 

2. Since the sun appears to move from east to west, when it is exactly 12 
o? clock at one place, it will be past 12 o'clock at all places east, and before 12 at 
all places west. Hence, if the difference of time between two places, be subtracted. 
from the time at the easterly place, the result will be the time at the westerly 
place ; and if the difference be added to the time at the westerly place the result 
will be the time at the easterly place. 

EXAMPLES FOR PRACTICE. 

1. When it is 9 o'clock at Washington, it is 8 h. 7 min. 4 sec. 
at St. Louis; the longitude of Washington being 77° V, west, 
what is the longitude of St. Louis ? Ans. 90° 15' west. 

2. The sun rises at Boston 1 h. 11 min. 56 sec. sooner than 
at New Orleans; the longitude of New Orleans being 89° 2' west, 



what is the longitude of Boston ? Ans. 71° 3' west. 



220 COMPOUND NUMBERS. 

3. When it is half past 2 o'clock in the morning at Havana, it 
is 7 h. 46 min. 40 sec. p. m. at the Cape of Good Hope ; the 
longitude of the latter place being 18° 28' east, what is the 
longitude of Havana ? Ans. 82° 22' west. 

4. The difference of time between Valparaiso and Rome is 6 h. 
8 min. 28 se#. ; what is the difference in longitude ? 

5. A gentleman traveling East from Fort Leavenworth, which 
is in 94° 44' west longitude, found, on arriving at Philadelphia, that 
his watch, an accurate time keeper, was 1 h. 18 min. 16 sec. slower 
than the time at Philadelphia ; what is the longitude of Philadel- 
phia? Ans. 75° 10' west. 

6. When it is 12 o'clock M. at San Francisco it is 3 h. 58 min. 
23 J sec. P. M. at Rochester, N. Y; the longitude of the latter 
place being 77° 51' W., what is the longitude of San Francisco? 

7. A gentleman traveling West from Quebec, which is in. 71° 
12' 15" W. longitude, finds, on his arrival at St. Joseph, that his 
watch is 2 h. 33 min. 53{| sec. faster than true time at the latter 
place. If his watch has kept accurate time, what is the longitude 
of St. Joseph ? Ans. 109° 40' 44" W. 

8. A ship's chronometer, set at Greenwich, points to 5 h. 40 

min. 20 sec. p. M., when the sun is on the meridian ; what is the 

ship's longitude ? Ans. 85° 5' west. 

Note 3. — Greenwich, Eng., is on the meridian of 0°, and from this meridian 
longitude is reckoned. 

9. The longitude of Stockholm being 18° 3' 30" E., when it is 
midnight there, it is 5 h. 51 min. 41| sec. P. M. at New York ; 
what is the longitude of New York from Greenwich ? 

Ans. 74° 1' 6" W. 

10. A vessel set sail from New York, and proceeded in a south- 
easterly direction for 24 days. The captain then took an obser- 
vation on the sun, and found the local time at the ship's meridian 
to be 10 h. 4 min. 36.8 sec. A. m. ) at the moment of the observa- 
tion, his chronometer, which had been set for New York time, 
showed 8 h. 53 min. 47 sec. Now, allowing that the chronometer 
had gained 3.56 sec. per day, how much had the ship changed her 
longitude since she set sail ? Ans. 15° 3' 48.6". 



LONGITUDE AND TIME. 221 



CASE II. 

386. To find the difference of time between two 
places or meridians, where the difference of longitude 
is known. 

Analysis. A difference of 15° of longitude produces a difference 
of 1 h. of time, 15' of longitude a difference of 1 min. of time, and 
15 y/ of longitude a difference of 1 sec. of time, (184). Hence the 

Rule. Divide the difference of longitude, expressed in degrees, 
minutes, and seconds, by 15, according to the rule for division of 
compound numbers; the quotient will be the difference of time in 
hours, minutes, and seconds. 

EXAMPLES FOR PRACTICE. 

1. Washington is 77° 1' and Cincinnati is 84° 24' west longi- 
tude ; what is the difference of time ? Ans. 29 min. 32 sec. 

2. Paris is 2° 20' and Canton 113° 14' east longitude; what is 
the difference in time ? 

3. Buffalo is 78° 55' west, and the city of Rome 20° 30' east 
longitude ; what is the difference in time ? 

Ans. 6 h. 37 min. 40 sec. 

4. A steamer arrives at Halifax, 63° 36' west, at 4 h. 30 min. 
p. m. ; the fact is telegraphed to New York, 74° 1' west, without 
loss of time ; what is the time of its receipt at New York ? 

5. The longitude of Cambridge, Mass., is 71° 7' west, and of 
Cambridge, England, is 5' 2" east; what time is it at the former 
place when it is 12 M. at the latter ? 

Ans. 7 h. 15 min. Hyf sec. A. M. 

6. The longitude of Pekin is 118° east, and of Sacramento 
City 120° west; what is the difference in time? 

7. The longitude of Jerusalem is 35° 32' east, and that of 
Baltimore 76° 37' west; when it is 40 minutes past 6 o'clock 
A. m. at Baltimore, what is the time at Jerusalem ? 

8. What time is it in Baltimore when it is 6 o'clock p. m. at 
Jerusalem? Ans. 10 h. 31 min. 24 sec. A.M. 

19* 



222 COMPOUND NUMBERS. 

9. The longitude of Springfield, Mass., is 72° 35' 45" W., and 
of Galveston, Texas, 94° 46' 34" W.j when it is 20 min. past 6 
o'clock A. m. at Springfield, what time is it at Galveston ? 

10. The longitude of Constantinople is 28° 49' east, and of St. 
Paul 93° 5' west; when it is 3 o'clock p. m. at the latter place, 
what time is it at the former ? 

11. What time is it at St. Paul when it is midnight at Constan- 
tinople? Ans. 3 h. 52 min. 24 sec. P. M. 

12. The longitude of Cambridge, Eng., is 5' 2" E., and of 
Mobile, Ala., 88° 1' 29" W.; when it is, 12 o'clock m. at Mobile, 
what is the time at Cambridge ? 

MISCELLANEOUS EXAMPLES IN COMPOUND NUMBERS. 

1. In 9 lb. 8g lj 29 19 gr. how many grains? 

2. How much will 3 cwt. 12 lb. of hay cost, at $15^ a ton? 

3. In 27 yd. 2 qr. how many Eng. ells? Ans. 22. 

4. Reduce $18,945 to sterling money. Ans. £3 12s. 6d. 

5. In 4 yr. 48 da. 10 h. 45 sec. how many seconds ? 

6. How many printed pages, 2 pages to each leaf, will there be 
in an octavo book having 24 fully printed sheets ? Ans. 384. 

7. At 1/6 sterling per yard, how many yards of cloth may be 
bought for £5 6s. 6d. ? Ans. 71 yd. 

8. In 4 mi. 51 ch. 73 1. how many links ? 

9. In 22 A. 3 R. 33 sq. rd. 2| sq. yd. how many square yards ? 

10. How many demijohns, each containing 3 gal. 1 qt. 1 pt., 
can be filled from 3 hhd. of currant wine ? Ans. 56. 

11. Paid $375.75 for 2^ tons of cheese, and retailed it at 9^ 
cts. a pound ; how much was my whole gain ? 

12. A gentleman sent a silver tray and pitcher, weighing 3 lb. 
9 oz., to a jeweler, and ordered them made into tea spoons, each 
weighing 1 oz. 5 pwt. ; how many spoons ought he to receive ? 

Ans. 3 doz. 

13. What part of 4 gal. 3 qt. is 2 qt. 1 pt. 2 gi. ? Ans. %fr 

14. Reduce | of T 4 T of a rod to the fraction of yard. 

15. How many yards of carpeting 1 yd. wide, will be required 
to cover a floor 26 J ft. long, and 20 ft. wide ? Ans. 57|. 



MISCELLANEOUS EXAMPLES. 223 

16. If I purchase 15 T. 3 cwt. 3 qr. 24 lb. of English iron, by 
long ton weight, at 6 cents a pound, and sell the same at $140 per 
short ton, how much will I gain by the transaction ? 

17. What will be the expense of plastering a room 40 ft. long, 
36} ft. wide, and 22i ft. high, at 18 cents a sq. yd., allowing 1375 
sq. ft. for doors, windows, and base board? Arts. $69.78}. 

18. How much tea in 23 chests, each weighing 78 lb. 9 oz. ? 

19. Valparaiso is in latitude 33° 2' south, and Mobile 30° 41' 
north ; what is their difference of latitude ? Ans. 63° 43'. 

20. If a druggist sell 1 gross 4 dozen bottles of Congress water 
a day, how many will he sell during the month of July ? 

21. Eighteen buildings are erected on an acre of ground, each 
occupying, on an average, 4 sq. rd. 120 sq. ft. 84 sq. in. ; how 
much ground remains unoccupied ? 

22. At $13 per ton, how much hay may be bought for $12.02} ? 

23. If 1 pk. 4 qt. of wheat cost $.72, how much will a bushel 
cost? Ans. $1.92. 

24. How many bushels, Indiana standard, in 36244 lbs. of 
wheat ? 

25. At 20 cents a cubic yard, how much will it cost to dig a 
cellar 32 ft. long, 24 ft. wide, and 6 ft. high? Ans. $34.13+. 

26. If the wall of the same cellar be laid 1} feet thick, what 
will it cost at $1.25 a perch ? Arts. 50.90{£. 

27. The forward wheels of a wagon are 15} ft. in circumfer- 
ence, and the hind wheels 10 ft. 4 in. ; how many more times will 
the forward wheels revolve than the hind wheels in running from 
Boston to N. Y., the distance being 248 miles ? Ans. 42240. 

28. Bought 15 cwt. 22 lb. of rice at $3.75 a cwt., and 7 cwt. 
36 lb. of pearl barley at $4.25 a cwt. How much would be gained 
by selling the whole at 4} cents a pound ? Ans. $13,255. 

29. From f of 3 T. 10 cwt. subtract -^ of 7 T. 3 cwt. 26 lb. 

30. What is the value in avoirdupois weight of 16 lb. 5 oz. 10 
pwt. 13 gr. Troy? Ans. 13 lb. 8 oz. 11.4+dr. 

31. What decimal of a rod is 1 ft. 7.8 in. ? 

32. If a piece of timber be 9 in. wide and 6 in. thick, what 
length of it will be required to make 3 cu. ft. ? Ans, 8 ft. 



224 COMPOUND NUMBERS. 

33. If a board be 16 in. broad, what length of it will make 7 
sq. ft. ? Ans. 5i ft. 

34. If a hogshead contain 30 cubic feet, how many more gal- 
lons of dry measure will it contain than of beer measure ? 

35. How many tons in a stick of hewn timber 60 ft. long, and 1 
ft. 9 in. by 1 ft. 1 in. ? Ans. 2.275 tons. 

36. Subtract ^ bu. + I of f § of 34 qt. from 5 bu. 3f ^ qt. 

Ans. 16f pk. 

37. What is the difference between I of 5 sq. mi. 250 A. 3 R., 
and 31 times 456 A. 3 R. 14 P. 25 sq. yd. ? 

Ans. 2 sq. mi. 254 A. 2 R. 26 P. 24| sq. yd. 

38. How many pounds of silver, Troy weight, are equivalent 
in value to 5.6 lb. of gold by the English government standard ? 

Ans. 80 lb. 2 pwt. 19.2768 gr. 

39. If a piece of gold is f pure, how many carats fine is it? 

40. In gold 16 carats fine what part is pure, and what part is 
alloy ? 

41. A man having a piece of land containing 384| A., divided 
it between his two sons, giving to the elder 22 A. 1 R. 20 P. more 
than to the younger \ how many acres did he give to each ? 

Ans. 203 A. 2 R. 14 P., elder ; 181 A. R. 34 P., younger. 

42. 4000 bushels of corn in Illinois is equal to how many bushels 
in New York ? Ans. 413|| bu. 

43. The market value being the same in both States, a farmer 
in New Jersey exchanged 110 bu. of cloverseed, worth $4 a 
bushel, with a farmer in New York for corn, worth $f a bushel, 
which he sold in his own State for cash. The exchange being 
made by weight, in whose favor was the difference, and how much 
in cash value ? 

Ans. The N. J. farmer gained 25^ bu. corn, worth $37|. 

44. The great pyramid of Cheops measures 763.4 feet on each 
side of its base, which is square. How many acres does it cover ? 

45. The roof of a house is 42 ft. long, and each side 20 ft. 6 
in. wide; what will the roofing cost at $4. 62 J a square ? 



' 



MISCELLANEOUS EXAMPLES. 225 

46. If 17 T. 15 cwt. 62* lb. of iron cost $1333.593, how much 
will 1 ton cost? 

47. How many wine gallons will a tank hold, that is 4 ft. long 
by 3f ft. wide, and If ft. deep? Ans. 187^ gal. 

48. What will be the cost of 300 bushels of wheat at 9s. 4d. 
per bushel, Michigan currency ? Ans. $350. 

49. What will be the cost in Missouri currency? 

50. What will be the cost in Delaware currency ? 3 '< 

51. What will be the cost in Georgia currency? Ans. $600. 

52. What will be the cost in Canada currency? Ans. $560. 

53. Bought the following bill of goods in Boston : 

6J yd. Irish linen @ 5/4 
12 " flannel " 3/9 

8i " calico " 1/7 

9 " ribbon " /9 

4i lb. coffee " 1/5 

6f gal. molasses " 3/8 

What was the amount of the bill ? Ans. $21.76 +. 

54. How many pipes of Madeira are equal to 22 pipes of 
sherry ? • 

55. A cubic foot of distilled water weighs 1000 ounces avoirdu- 
pois ; what is the weight of a wine gallon ? Ans. 8 lb. 5^| oz. 

56. There is a house 45 feet long, and each of the two sides 
of the roof is 22 feet wide. Allowing each shingle to be 4 inches 
wide and 15 inches long, and to lie one third to the weather, how 
many half-thousand bunches will be required to cover the roof? 

Ans. 342-A&. 

57. A cistern measures 4 ft. 6 in. square, and 6 ft. deep; how 
many hogsheads of water will it hold ? 

58. If the driving wheels of a locomotive be 18 ft. 9 in. in cir- 
cumference, and make 3 revolutions in a second, how long will the 
locomotive be in running 150 miles ? 

Ans. 3 h. 54 min. 40 sec. 
59 In traveling, when I arrived at Louisville my watch, which 
was exactly right at the beginning of my journey, and a correct 

P 



226 COMPOUND NUMBERS. 

timekeeper, was 1 h. 6 min. 52 sec. fast; from what direction 
had I come, and how far? Ans. From the east, 16° 43'. 

60. How many U. S. bushels will a bin contain that is 8.5 ft. 
long, 4.25 ft. wide, and 3f ft. deep? 

61. Reduce 3 hhd. 9 gal. 3 qt. wine measure to Imperial gal- 
lons. Ans. 165.5804 Imp'l gal. 

62. A man owns a piece of land which is 105 ch. 85 1. long, 
and 40 ch. 15 1. wide; how many acres does it contain ? 

63. A and B own a farm together; A owns T 7 2 of it and B the 
remainder, and the difference between their shares is 15 A. 1 It. 
28 J P. How much is B's share ? Ans. 38 A. 2 R. Ill P. 

64. At $3.40 per square, what will be the cost of tinning both 
sides of a roof 40 ft. in length, and whose rafters are 20 ft. 6 in. 
long? Ans. $55.76. 

65. What is the value of a farm 189.5 rd. long and 150 rd. 
wide, at $3 If per acre ? 

66. Reduce 9.75 tons of hewn timber to feet, board measure, 
that is, 1 inch thick. Ans. 5850 ft. 

67. How many wine gallons will a tank contain that is 4 ft. 
long, %l ft. wide, and 2f ft. deep? Ans. 2991? gal. 

68. If a load of wood be 12 ft. long, and 3 ft. 6 in. wide, how 
high must it be to make a cord ? 

69. In a school room 32 ft. long, 18 ft. wide, and 12 ft. 6 in. 
high, are 60 pupils, each breathing 10 cu. ft. of air in a minute. 
In how long a time will they breathe as much air as the room 
contains ? 

70. A man has a piece of land 201f rods long and 41 J rods 
wide, which he wishes to lay out into square lots of the greatest 
possible size. How many lots will there be ? Ans. 396. 

71. A man has 4 pieces of land containing 4 A. 3 R. 20 P., 6 
A. 3 R. 12 P., 9 A. 3 R., and 11 A. 2 R. 32 P. respectively. 
It is required to divide the whole into the largest sized building 
lots possible, each lot containing the same area, and an exact num- 
ber of square rods. How much land will each lot contain ? 

Ans. 156 P. 



DUODECIMALS. 227 



DUODECIMALS. 

387. Duodecimals are the parts of a unit resulting from con- 
tinually dividing by 12 ; as 1, j\, T | ? , ttW etc - ^ n practice, 
duodecimals are applied to the measurement of extension, the foot 
being taken as the unit. 

In the duodecimal divisions of a foot, the different orders of 
units are related as follows : 

Y (inch or prime) is J^ of a foot, or 1 in. linear measure. 

V (second) or ^ of ^ , " T J ¥ of a foot, or 1 ** square " 

1 7// (third) or T 1 ^ of t l of fo . . " jfa of a foot, or 1 " cubic « 

TABLE. 

12 fourths, ( r/// ), make 1 third 1"' 

12 thirds u 1 second, 1" 

12 seconds ** 1 prime, or inch, . 1' 

12 primes, or inches, " 1 foot, ft. 

Scale — uniformly 12. 

The marks ', ", //V , /y// , are called indices. 

Notes. — I. Duodecimals are really common fractions, and can always be 
treated as such ; but usually their denominators are not expressed, and they are 
treated as compound numbers. 

2. The word duodecimal is derived from the Latin term duodecim, signifying 
12. 

ADDITION AND SUBTRACTION. 

388. Duodecimals are added and subtracted in the same 
manner as compound numbers. 

EXAMPLES FOR PRACTICE. 

1. Add 12 ft. T 8", 15 ft. 3' 5", 17 ft. 9' 7". 

Ans. 45 ft. 8' 8". 

2. Add 136 ft. 11' 6" 8'", 145 ft. W 8" 5'", 160 ft, 9' 5" 5"'. 

Ans, 443 ft. T 8" &". 

3. From 36 ft. 7 11" take 12 ft. 9' 5". Ans, 23 ft. 10' 6". 

4. A certain room required 300 sq. yd. 2 sq. ft 5 f of plastering. 
The walls required 50 sq. yd. 1 sq. ft. T 4", 62 sq. yd. 5' 3", 48 
sq. yd. 2 sq. ft., and 42 sq. yd. 2 sq. ft. 3' 4", respectively. Re- 
quired the area of the ceiling. Ans. 97 sq. yd. 5 sq. ft. V 1". 



OPERATION. 


9 ft. 


8' 




4 ft. 


7' 




5 ft. 


7' 


8" 


38 ft. 


8' 





228 DUODECIMALS. 

MULTIPLICATION. 

389. In the multiplication of duodecimals, the product of two 
dimensions is area, and the product of three dimensions is 
solidity (282, 386). 

We observe that 

V xlft. = 1 > 5 of lft. =1'. 

1"X 1 ft. = t^t of lft. = 1". 

y x v = y* x T >2 of l ft. = l". 

1// x 1' = T h X 3V of 1 ft. = 1"'. Hence, 
The product of any two orders is of the order denoted by the 
sum of their indices. 

390. 1. Multiply 9 ft. 8' by 4 ft. 7'. 

Analysis. Beginning at the right, 

8' X V = 56" = 4' 8" ; writing 
the 8" one place to the right, we re- 
serve the 4' to be added to the next 
product. Then, 9 ft. X V + ¥.= 
67 7 = 5 ft. 7', which we write in the 
44 ft. 3' 8", Ans. places of feet and primes. Next mul- 

tiplying by 4 ft., we have 8' X 4 ft. 
= 32' = 2 ft. 8' ; writing the 8 / in the place of primes, we reserve the 
2 ft. to be added to the next product. Then, 9 ft. X 4 ft. + 2 ft. = 
38 ft., which we write in the place of feet. Adding the partial pro- 
ducts, we have 44 ft. 3 / 8" for the product required. Hence the 

Rule. I. Write the several terms of the multiplier under the 
corresponding terms of the multiplicand, 

II. Multiply each term of the multiplicand by each term of the 
multiplier, beginning with the lowest term in each, and cull the pro- 
duct of any two orders, the order denoted by the sum of their in- 
dices, carrying 1 for every 12. 

III. Add the partial products ; their sum will be the required 
answer. 

EXAMPLES FOR PRACTICE. 

1. How many square feet in a floor 16 ft. 8' wide, and 18 ft. 5' 
long? 

2. How much wood in a pile 4 ft. wide, 3 ft. 8' high, and 23 ft. 
7' long? 



MULTIPLICATION. 229 

3. If a floor be 79 ft. 8' by 38 ft. 11', how many square yards 
does it contain ? Ans. 344 yd. 4 ft. 4' 4". 

4. If a block of marble be 7 ft. 6' long, 3 ft. 3' wide, and 1 ft. 
10' thick, what are the solid contents? Ans. 44 ft. 8' 3". 

5. How many solid feet in 7 sticks of timber, each 56 ft. long, 
11 inches wide, and 10 inches thick? Ans. 299 ft. 5' 4". 

6. How many feet of boards will it require to inclose a building 
60 ft. 6' long, 40 ft. 3' wide, 22 ft. high, and each side of the 
roof 24 ft. 2', allowing 523 ft. 3' for the gables, and making no 
deduction for doors and windows ? Ans. 7880 ft. 5'. 

CONTRACTED METHOD. 

391. The method of contracting the multiplication of deci- 
mals may be applied to duodecimals, the only modification being 
in carrying according to the duodecimal, instead of the decimal, 
scale. 

1. Multiply 7 ft. 3' 5" 8'" by 2 ft. 4' 7" 9'", rejecting all de- 
nominations below seconds in the product. 

Analysis. We write 2 ft., the 
units of the multiplier, under the 
lowest order to be reversed in the 
product, and the other terms at the 
left, with their order reversed. Then 
it is obvious that the product of 
each term by the one above it is 
17~ft 4/ jy'-h Jji S seconds. Hence we multiply each 
term of the multiplier into the terms 
above and to the left of it in the multiplicand, carrying from the 
rejected terms, thus ; in multiplying by 2 ft., we have 8"' X 2 ft. = 
16 //7 = \" 4 //7 , which being nearer V than 2", gives 1" to be car- 
ried to the first contracted product. In multiplying by 4 / , we have 
5" x 4' = 20 /7/ = 1" 8'", which being nearer 2" than 1 // , gives 
2" to be carried to the second contracted product, and so on. 

EXAMPLES FOR PRACTICE. 

1. Multiply 7 ft. 3' 4" 5'" by 5 ft. 8" 6", extending the pro- 
duct only to primes. Ans. 41 ft. 7'=k. 
20 



OPERATION. 


7 ft. 3' 


5" 8'" 


9'" 7" 


4' 


2ft 


14 ft. 


6' 


11" 


2 ft. 


5' 


2" 




4' 


8" 

5" 



4 ft. 7' 
4 ft. 3' 


7" 
11" 


3' 
3' 


.8" 6'" 
8" 6'" 



230 DUODECIMALS. 

2. How many yards of carpeting will cover a floor 36 ft. 9' 4" 
long, and 26 ft. 6' 9" wide ? 

3. How many cu. ft. in a block of marble measuring 6 ft. 2! 7" 
in length, 3 ft. 3' 4" wide, and 2 ft. 8' 6" thick? 

4. Find the product of 7 ft. 6' 8", 3 ft. 2' 11", and 3 ft. 8' 4", 
correct to within 1'. Arts. 90 ft. 2'±. 

DIVISION. 

392. 1- Divide 41 ft. 8' 7" 6'" by 7 ft. 5'. 

operation. Analysis. Divid- 

7 ft. 5')41 ft. 8' 7" 6"'(5 ft. T 6" ing the units of the 

37 ft. 1' dividend by the units 

of the divisor, we ob- 
tain 5 ft. for the first 
term of the quotient, 
and 4 ft. 7' for a re- 
mainder. Bringing 
down the next term of the dividend, we have 4 ft. 7 / 7" for a new 
dividend. Eeducing the first two terms to seconds, we have 55' 7 // , 
whence by trial division we obtain 7 / for the second term of the quo- 
tient, and y 8" for a remainder. Completing the division in like 
manner, we have 5 ft. 7 / 6" for the entire quotient. Hence the fol- 
lowing 

Rule. I. Write the divisor on the left hand of the dividend, 
as in simple numbers 

II. Find the first term of the quotient either by dividing the 
first term of the dividend by the first term of the divisor, or by 
dividing the first two terms of the dividend by the first two terms 
of the divisor ; multiply the divisor by this term of the quotient, 
subtract the product from the corresponding terms of the dividend, 
and to the remainder bring down another term of the dividend. 

III. Proceed in like manner till there is no remainder, or till a 
quotient has been obtained sufficiently exact. 

EXAMPLES FOR PRACTICE. 

1. Divide 287 ft. T by 17 ft. Ans. 16 ft. 11'. 

2. Divide 29 ft. 5' 4" by 6 ft. 8'. Ans. 4 ft. 5'. 



DIVISION. 231 

3. A floor whose length is 48 ft. 6' has an area of 1176 ft. 1' 
6"; what is its width ? Ans. 24 ft. 3'. 

4. From a cellar 38 ft. 10' long and 9 ft. 4' deep, were exca- 
vated 275 cu. yd. 5 cu. ft. V 4" of earth; how wide was the 
cellar ? Ans. 20 ft, 6'. 

CONTRACTED METHOD. 

393. Division of Duodecimals may be abbreviated after the 
manner of contracted division of decimals. 

1. Divide 35 ft. 11' 11" by 4 ft. 3' 7" 3"' ? and find a quotient 
correct to seconds. 

OPERATION. 

4 ft. 3' 7" 3'" ) 35 ft. 11' 11" ( 8 ft. 4' 5" 
34 ft. 4' 10" 



lft. 


7' 


i" 


1ft. 


5' 


2" 




1' 


11" 




r 


9" 



2", rem. 

Analysis. Having obtained by trial, 8 ft. for the first term of the 
quotient, we multiply three terms of the divisor, 4 ft. 3' 7", carrying 
from the rejected term, 3 7// X 8 = 24 /// = 2 // , making 34 ft. 4' 10", 
which subtracted from the dividend leaves 1 ft. 7 / V for a new divi- 
dend. In the next division, we reject 2 terms from the right of the 
divisor, and at the last division, 3 terms, and obtain for the required 
quotient, 8 ft. 4' 5". 

EXAMPLES FOR PRACTICE. 

1. Divide 7 ft. 7' 3" by 2 ft. 10' 7", extending the quotient to 
thirds. Ans. 2 ft. 7' 8"±. 

2. Separate 64 ft. 9' 8" into three factors, the first and second 
of which shall be 7 ft. 2' 4" and 4 ft. 7' 9" 8'" respectively, and 
obtain the third factor correct to within 1 second. 

Ans. 1 ft. 11' 2"=±r. 

3. What is the width of a room whose area is 36 ft. 4' 8" and 
whose length 7 ft. 2' 11" ? 



232 SHORT METHODS 

SHORT METHODS. 

1194. Under the heads of Contractions in Multiplication and 
Contractions in Division, are presented only such short methods 
as are of the most extensive application. The short methods 
which follow, although limited in their application, are of much 
value in computations. 

FOR SUBTRACTION. 

395. "When the minuend consists of one or more 
digits of any order higher than the highest order in the 
subtrahend. 

The difference between any number and a unit of the next 
higher order is called an Arithmetical Complement. Thus, 4 is the 
arithmetical complement of 6, 31 of 69, 2792 of 7208, etc. 

I. Subtract 29876 from 400000. 

operation. Analysis. To subtract 29876 from 400000 is the 

400000 same as to subtract a number one less than 29876, or 

29876 29875, from 399999 (Ax. 2). We therefore diminish 
' 07AIQ4 the 4 of the minuend by 1, and then take each figure 
of the subtrahend from 9, except the last or right- 
hand digit, which we subtract from 10. Hence the 

Rule. I. Subtract 1 from the digit of the minuend and write 
the remainder, if 'any ', as the first figure in the result. 

II. Commencing at the left, subtract each figure in the subtra- 
hend from 9, except the last significant figure, which subtract from 
10. 

EXAMPLES FOR PRACTICE. 

1. Subtract 756 from 1000. Ans. 244. 

2. Subtract 8576 from 4000000. Ans. 3991424. 

3. Subtract .5768 from 10. 

4. Subtract 13057 from 1700000. 

5. Subtract 90.59876 from 64000. 

6. Subtract 599948 from 1000000. 

7. What is the arithmetical complement of 271 ? Of 18365 ? 
Of 3101250? 



FOR MULTIPLICATION. 23$ 

FOR MULTIPLICATION. 
CASE I. 

396. When the multiplier is 9, 99, or any number 

of 9's. 

Annexing 1 cipher to a number multiplies it by 10, two ciphers by 
100, three ciphers by 1000, etc. Since 9 is 10 — 1, any number may 
be multiplied by 9 by annexing 1 cipher to it and subtracting the 
number from the result. For similar reasons, 100 times a number — 
1 time the number = 99 times the number, etc. Hence, 

Rule. Annex to the number as many ciphers as the number 
contains 9's, and subtract the number from the result. 

EXAMPLES FOR PRACTICE. 

1. Multiply 784 by 99. Ans. 77616. 

2. Multiply 5873 by .999. 

3. Multiply 4783 by 99999. Ans. 478295217. 

4. Multiply 75 by 999.999. 

CASE II. 

397. When the multiplier is a number a few units 
less than the next higher unit. 

Were we required to multiply by 97, which is 100 — 3, we could 
evidently annex 2 ciphers to the multiplicand, and subtract 3 times 
the multiplicand from the result. Were our multiplier 991, which is 
1000 — 9, we could subtract 9 times the multiplicand from 1000 times 
the multiplicand. Hence, 

Rule. I. Multiply by the next higher unit by annexing 
ciphers. 

II. From this result subtract as many times the multiplicand 
as there are units in the difference between the multiplier and the 
next higher unit. 

EXAMPLES FOR PRACTICE. 

1. Multiply 786 by 98. Ans. 77028. 

2. Multiply 4327 by 96. Ans. 415392. 

3. Multiply 7328 by 997. 

20* 



234 SHORT METHODS 

4. Multiply 7873.586 by 9.95. Ans. 78242.18070. 

5. Multiply 43789 by 9994. 

6. Multiply 7077364 by .999993. 

CASE III. 

398. When the left hand figure of the multiplier is 
the unit, 1, the right hand figure is any digit whatever, 
and the intervening figures, if any, are ciphers. 

I. Multiply 3684 by 17. 

operation. Analysis. If we multiply by the usual 

3684 X 17 method, we obtain, separately, 7 times and, 

ayanQ 10 times the multiplicand, and add them. 

We may therefore multiply by the 7 units, 

and to the product add the multiplicand regarded as tens, thus : 7 times 

4 is 28, and we write the 8 as the unit figure of the product. Then, 

7 times 8 is 56, and the 2 reserved being added is 58, and the 4 in 

the multiplicand, added, is 62, and we write 2 in the product. Next, 

7 times 6, plus the 6 reserved, plus the 8 in the multiplicand, is 56, 

and we write 6 in the product. Next, 7 times 3, plus the 5 reserved, 

plus the 36 in the multiplicand, is 62, which we write in the product, 

and the work is done. 

Had the multiplier been 107, we should have multiplied two figures 
of the multiplicand by 7, before we commenced adding the digits of 
the multiplicand to the partial products ; 3 figures had the multiplier 
been 1007, etc. Hence the 

Rule. I. Write the multiplier at the right of the multiplicand, 
with the sign of multiplication between them. 

II. Multiply the multiplicand by the unit figure of the multi- 
plier, and to the product add the multiplicand, regarding its local 
value as a product by the left hand figure of the multiplier. 

EXAMPLES FOR PRACTICE. 

1. Multiply 567 by 13. Ans. 7371. 

2. Multiply 439603 by 10.5. Ans. 4615831.5. 

3. Multiply 7859 by 107. 

4. Multiply 18075 by 1008. Ans. 18219600. 
5'. Multiply 3907 by 10.002. 



FOR MULTIPLICATION. 235 

CASE IV. 

399. When the left hand figure of the multiplier is 
any digit, the right hand figure is the unit, 1, and the 
intermediate figures, if any, are ciphers. 

I. Multiply 834267 by 301. 

operation. Analysis. Regarding the multipli- 

834267 X 301 can( * as a product by the unit, 1, of the 

multiplier, we multiply the multipli- 

251114db7 cand by 3 hundreds, and add the digits 

of the multiplicand to the several products as we proceed. Since the 

3 is hundreds, the two right hand figures of the multiplicand will 

be the two right hand figures of the product ; and the product of 3 X 7 

will be increased by 2, the hundreds of the multiplicand. 

Had the multiplier been 31, the tens of the multiplicand would 
have been added to 3 X 7 ; had the multiplier been 3001 the thousands 
of the multiplicand would have been added to 3 X 7 ; and so on. 
Hence the 

Rule. I. Write the multiplier at the right of the multiplicand, 
with the sign of multiplication between them. 

II. Multiply the multiplicand by the left hand figure of the mul- 
tiplier, and to the product add the multiplicand, regarding its local 
value as a product by the unit figure of the multiplier. 

EXAMPLES FOR PRACTICE. 

1. Multiply 56783 by 71.. 

2. Multiply 47.89 by 60.1. Ans. 2878.189. 

3. Multiply 3724.5 by .901, 

4. Multiply 103078 by 10004. Ans. 4123223078. 

CASE V. 

400. When the digits of the multiplier are all the 
same figure. 

1. Multiply 81362 by 333. 

operation. Analysis. We first multiply by 999, by 

81362000 (396). Then, since 333 is J of 999, we take 

81362 i of the product. 

o n Qi9Qf)fi38 •^■ a( * our mu ltipli er been 444, we would 

- have taken £ of 999 times the multiplicand. 

27093546 Had it been ffi, we would have taken £ == f 
of 99 times the multiplicand, etc. Hence 



236 SHOUT METHODS 

Rule I. Multiply by as many 9's as the multiplier contains 
digits, by (396). 

II. Take such a part of the product as 1 digit of the multiplier 
is part of 9. 

EXAMPLES FOR PRACTICE. 

1. Multiply 432711 by 222. Ans. 96061842. 

2. Multiply 578 by 1111. 

3. Multiply .6732 by 88.888. Ans. 59.8394016. 

4. Multiply 8675 by 77.7. 

5. Multiply 44444 by 88888. 

CASE VI. 

401. To square a number consisting of only two 
digits. 

I. What is the square of 18 ? 
Analysis. According to (86), we have 

18 2 =18x 18 
Now if one of these factors be diminished by 2, the product will be 
less than the square of 18 by 2 times the other factor, (93, 1) ; that is, 
182= (16 x 18) + (2x18). 
Next, if we increase the other factor, 18, in this result, by 2, the 
whole result will:exceed the square of 18, by 2 times the other factor, 
16, (93, II) ; that is, 

18 2 = (16 X 20) + (2 X 18) — (2 X 16). 
But as 2 times 18 minus 2 times 16 is equal to 2 X 2, or 2 2 , we 

have 

18 2 = 16 X 20 + 2 2 . Hence the 

Rule. I. Take two numbers, one of which is as many units 
less than the number to be squared as the other is units greater, and 
one of the numbers taken an exact number of tens. 

II. Multiply these two numbers together, and to the product add 
the square of the difference between the given and one of the as- 
sumed numbers. 

Note. — A little practice will enable the pupil to readily square any number 
leas than 100 mentally by this rule. 



FOR MULTIPLICATION. 237 

EXAMPLES FOR PRACTICE. 

1. What is the square of 27 ? Ans. 729. 

2. What is the square of 49 ? Ajis. 2401. 

3. Square 28, 26, 39, 38, 37, 36, and 35. 

4. Square 77, 88, 8.6, 99, 98, 69, 68, 6.7, and 62. 

CASE VII. 

402. WTien the multiplier is an aliquot part of some 
higher unit. 

An Aliquot or Even Part of a number is such a part as will 

exactly divide that number. Thus, 5, 8 J, and 12 J are aliquot 

parts of 25 and of 100, etc. 

Note. — An aliquot part may be either a whole or a mixed number, while a 
component factor must be a whole number. 

403. The aliquot parts of 10 are 5, 31, 2J, 2, If, If, 1|, If 

The aliquot parts of 100, 1000, or of any other number, may 
be found by dividing the number by 1, 2, 3, 4, etc., until it has 
been divided by all the integral numbers between 1 and itself. 

I. Multiply 78 by 3J, and by 25 separately. 

operation. Analysis. Since 3J is J of 10, 

3 ) 780 4 ) 7800 the next higher unit, we multiply 

2(30 a 1950 ^ kv 10 and take i of the product. 

Again, since 25 is J of 100, we 

multiply 78 by 100 and take J of the product. Hence the 

Rule. I. Multiply the given multiplicand by the unit next 
higher than the multiplier, hy annexing ciphers. 

II. Take such a part of this product as the given multiplier is 
part of the next higher unit. 

EXAMPLES FOR PRACTICE. 



1. 


Multiply 437 by 25. 


Ans. 10925. 


2. 


Multiply 6872 by 2|. 


Ans. 17180. 


3. 


Multiply 5734154 by 333f 


Ans. 1911384666|. 


4. 


Multiply 758642 by 12^. 




5. 


Multiply 78563 by 125. 


Ans. 9820375. 


6. 


Multiply 57687 by 142f . 





238 



SHORT METHODS 



CASE VIII. 

404. When the right hand figure or figures of the 
multiplier are aliquot parts of 10, 100, 1000, etc. 

I. Multiply 2183 by 1233J. 

OPERATION. 

218300 

12 J Analysis. 1233J = 12J X 100. We there- 

797ffi 2 *° re mu ltiply ^y 100, and by 12J, in continued 

o pi of* multiplication. Hence the 

26923661 
Rule. I. Reject from the right hand of the multiplier such 
figure or figures as are an aliquot part of some higher unit, and 
to the remaining figures of the multiplier annex a fraction which 
expresses the aliquot part thus rejected, for a reserved multiplier. 

II. Annex to the multiplicand as many ciphers as are equal to 
the number of figures rejected from the right hand of the multi- 
plier, and multiply the result by the reserved multiplier. 

EXAMPLES FOR PRACTICE. 

1. Multiply 43789 by 825. Ans. 36125925. 

2. Multiply 58730 by 7125. 

3. Multiply 7854 by 34.2J. Ans. 268999.5. 

4. Multiply 30724 by 73333 J. 

5. Multiply 47836 by 712J. Ans. 34083150. 

6. Multiply 53727 by 24161 . 

CASE IX. 

405. To find the cost of a quantity when the price 
is an aliquot part of a dollar. 

1. What cost a case of muslins containing 1627 yds., at $.12 \ 
per yard ? 

operation. Analysis. At $1 per yard the case would 

8 ) $1627 cost as many dollars as it contained yards ; 

S203 37£ an( * at $-1^ == $£ P er y ar( *> ** would cost J 

as many dollars as it contained yards. We 

therefore regard the yards as dollars, which we divide by 8. Hence, 



FOR MULTIPLICATION. 239 

Rule. Take such a part of the given quantity as the price is 

part of one dollar. 

Note. — Since the shilling in most of the different currencies is some aliquot 
part of the dollar, this rule is of much practical use in making out bills and 
accounts where the prices of the items are given in State Currency, and the 
amounts are required in United States Money. 

EXAMPLES FOR PRACTICE. 

1 . What cost 568 pounds of butter at 25 cents a pound ? 

Ans. 8142. 

2. A merchant sold 51 yards of prints at 16f cents per yard, 8 
pieces of sheeting, each piece containing 33 yards, at 6i cents per 
yard, and received in payment 18 bushels of oats at 33 i cents per 
bushel, and the balance in money ; how much money did he re- 
ceive ? Ans. $19. 

3. Required the cost of 28 dozen candles, at 1 shilling per 
dozen, New York currency. Ans. $3.50. 

4. "What cost 576 lbs. of beef at lOd. per pound, Pennsylvania 
currency? Ans. $64. 

5. If a grocer in New York gain $7,875 on a hogshead of mo- 
lasses containing 63 gallons, how much will he gain on 576 gallons 
at the same rate ? Ans. $72. 

CASE X. 

406. To find the cost of a quantity, when the quan- 
tity is a compound number, some part or all of which is 
an aliquot part of the unit of price. 

1. What cost 5 bu. 3 pk. 4 qt. of cloverseed, at $3.50 per bu. ? 

operation. Analysis. Multiply- 

2" 4" 8 ) $3.50 price. ing the price by 5, we 

5 have the cost of 5 bu. 

$17\50 cost of 5 bu. Dividing the price by 2, 

1 75 " " 2 Dk we nave the cost of J bu. 

.875 " « \ « =2 pk. Dividing the 

.4375 *' "4 qt. price by 4, or the cost of 

82075625, Ans. 2 P*' ^ 2 ' ™ ^^ 

cost of 1 pk. Dividing 

the price by 8, or the cost of 2 pk. by 4, or the cost of 1 pk. by 2, we 






240 SHORT METHODS 

have the cost of J pk. = 4 qt. And the sum of these several values 
is the entire cost required. 

2. At £6 7s. 5Jd. Sterling per hhd., how much will 4 hhd. 9 
gal. 3 qt. of West India Molasses cost ? 

operation. Analysis. Mul- 

7) £6 7s. 5Jd. price. tiplying the price 

4 by 4, we have the 

" 25 9 " 10 " cost of 4 hhd. cost of 4 hhd - Di " 

12) " 18 " 2 " 2 qr. " « 9 gal. viding the price by 

I « q a 5 u u u 3 q£ # 7, W e have the cost 

" 26 9« 6 " 2§, " ^s. of * hhd - = 9 S aL 

Dividing the cost of 

9 gal. by 12, we have the cost of -fe of 36 qt. = 3 qt. And the sum 

of these several results is the entire cost required. 

From these illustrations we deduce the following 

Rule. I. Multiply the price by the number of units of the de- 
nomination corresponding to the price. 

II. For the lower denominations, take aliquot parts of the price; 
the sum of the several results will be the entire cost. 

Note. — This method is applicable in certain cases of multiplication, where one 
compound number is taken as many times as there are units and parts of a 
unit of a certain kind, in another compound number. This will be seen in the 
first example below. • 

EXAMPLES FOR PRACTICE. 

1. A chemist filtered 18 gal. 3 qt. 1 pt. of rain-water in 1 day; 
at the same rate how much could he filter in 4 da. 6 h. 30 min. ? 
operation. Analysis. Multi- 

18 gal. 3 qt. 1 pt. in 1 da. P^g the quantity 

4 filtered in 1 day by 4, 

we have the quantity 
filtered in 4 days. Di- 
viding the quantity fil- 
tered in 1 day by 4, 
80 " 2 " " S T \ « Ans. we have the quantity 

filtered in J da. = 6 
h. Dividing the quantity filtered in 6 hours by 12, we have the quan- 
tity filtered in J h. = 30 min. And the sum of these several results 
is the entire result required. 



75 


a 


2 


U 





u 




it 


4 " 


4 


ti 


2 


a 


1 


a 


3 gi. 


u 


6 h. 






1 


u 


1 


u 


7 CC 


a 


30 min 



FOR DIVISION. 241 

2. What will be the cost of 3 lb. 10 oz. 8 pwt. 5} gr. of gold 
at $15.46 per oz. ? Ans. $717.51. 

3. A man bought 5 cwt. 90 lb. of hay at $.56 per cwt. ; what 
was the cost? Ans. $3,304. 

4. What must be given for 3 bu. 1 pk. 3 qt. of cloverseed, at 
$4.48 per bushel ? Ans. $14.98. 

5. A gallon of distilled water weighs 8 lb. 5 oz. 6.74 dr. ; re- 
quired the weight of 5 gal. 3 qt. 1 pt. 3 gi. 

Ans. 49 lb. 12 oz. 5.73— dr. 

6. At $17.50 an acre, what will 3 A. 1 E. 35.4 P. of land cost? 

7. If an ounce of English standard gold be worth <£31 7s. 10 Jd., 
what will be the value of an ingot weighing 7 oz. 16 pwt. 18 gr. ? 

Ans. £30 10s. 4.14375d. 

8. If a comet move through an arc of 4° 36' 40" in 1 day, how 
far will it move in 5 da. 15 h. 32 min. 55 sec. ? 

9. What will be the cost of 7 gal. 1 qt. 1 pt. 3 gi. of burning 
fluid, at 4s. 8d. per gallon, N. Y. currency? Ans. $4.35+. 

10. What must be paid for 12J days' labor, at 5s. 3d. per day, 
New England currency ? 

FOR DIVISION. 
CASE I. 

407. When the divisor is an aliquot part of some 
higher unit. 

1. Divide 260 by Si, and 1950 by 25. 

operation. Analysis. Since 3J is J of 10, the next 

26|0 19|50 higher unit, we divide 260 by 10 ; and hav- 

3 and 4 ing used 3 times our true divisor, we obtain 

^7j i^ only j- of our true quotient. Multiplying 

the result, 26, by 3, we have 78, the true 

quotient. Again, since 25 is £ of 100, the next higher unit, we divide 

1 1950 by 100 ; and having used 4 times our true divisor, the result, 

19.5, is only \ of our true quotient. Multiplying 19.5 by 4, we have 

78, the true quotient. Hence the 

Rule. I. Divide the given dividend by a unit of the order 
1 next higher than the divisor, by cutting off ^qures from the right. 
21 Q 






242 SHORT METHODS. 

II. Take as many times this quotient as the divisor is contained 
times in the next higher unit. 

EXAMPLES FOR PRACTICE. 

1. Divide 63475 by 25. A?is. 2539. 

2. Divide 7856 by 1.25. Ans. 6284.8. 

3. Divide 516 by 33.3J. 

4. Divide 16.7324 by 12J. 

5. Divide 1748 by .14f . Ans. 12236. 

6. Divide 576.34 by 1.6f. 

CASE II. 

408. When the right hand figure or figures of the 
divisor are an aliquot part of 10, 100, 1000, etc. 

1. Divide 26923661 by 1233J. 

Analysis. Since 33J is 

OPERATION. i * -inn n- 1 v xi~ 

i of 100, we multiply both 

1233 i ) 26923661 dividend and divisor by 3, 

(117, III), and we obtain a 



3 3 



37|00 ) 80771 100 ( 2183, Ans. divisor the component fac- 
67 tors of which are 100 and 37. 

307 We then divide after the 

H* manner of contracted divi- 

sion, (112). 
2. Divide 601387 by 1875. 

operation. Analysis. Multiplying both 

1875) 601387 dividend and divisor by 4, we ob- 

4 4 tain a new divisor, 7500, having % 

- CA n . TTaTcTo ciphers on the right of it. Multi- 

7500)2405548 *\ . , 7* .. . . 

a J , ply m g again by 4, we obtain a new 

divisor, 30000, having 4 ciphers on 



310000) 962|2192 the right. Then dividing the new 

320^112 Ans. dividend by the new divisor, we ob- 
tain 320 for a quotient, and 22192 
for a remainder. As this remainder is a part of the new dividend, 
it must be 4 X 4 = 16 times the true remainder ; we therefore divide 
it by 16, and write the result over the given divisor, 1875, and annex 
the fraction thus formed to the integers of the quotient. 



RATIO. 243 

Prom these illustrations we derive the following 

Rule. I. Multiply both dividend and divisor by a number or 

numbers that will 'produce for a new divisor a number ending in a 

cipher or ciphers. 

II. Divide the new dividend by the new divisor. 

Note. — If the divisor be a whole number, or a pure decimal, the multiplier 
will be 2, 4, 5, or 8, or some multiple of one of these numbers. 

EXAMPLES FOR PRACTICE. 

1. Divide 64375 by 2575. 

2. Divide 76394 by 3625. Arts. 21^%. 

3. Divide 7325 by 433£. 

4. Divide 5736 by 431.25. Ans. 13£Jf . 

5. Divide 42.75 by 5661. 

6. Divide 24409375 by .21875. 

7. Divide 785 by 3.14f . Ans. 249*|. 



RATIO. 

409. Ratio is the relation of two like numbers with respect 
to comparative value, expressed by dividing one by the other. 

Note. — There are two methods of comparing numbers with respect to value; 
1st, by subtracting one from the other; 2d, by dividing one by the other. The 
relation expressed by the difference is sometimes called Arithmetical Batio, and 
the relation expressed by the quotient, Geometrical Batio. 

4:10. When one number is compared with another, as 4 with 
I 12, by means of division, thus, 12 -j- 4 = 3, the quotient, 3, shows 
1 the relative value of the dividend when the divisor is considered 
j as a unit or standard. The ratio in this case shows that 12 is 3 
} times 4 ; that is, if 4 be regarded as a unit, 12 will be 3 units, or 
the relation of 4 to 12 is that of 1 to 3. 
411. Ratio is indicated in two ways : 

1st. By placing two points between the two numbers compared, 
writing the divisor before and the dividend after the points. 
Thus, the ratio of 8 to 24 is written 8 : 24 ; the ratio of 7 to 5 is 
written 7 : 5. 



1 



244 RATIO. 



2d. In the form of a fraction. Thus, the ratio of 8 to 24 is 
written \ 4 ; the ratio of 7 to 5 is f . 

413. The Terms of a ratio are the two numbers compared. 
The Antecedent is the first term; and 

The Consequent is the second term. 

The two terms of a ratio taken together are called a couplet 

41«$. A Simple Ratio consists of a single couplet; as 5 : 15. 

414. A Compound Ratio is the product of two or more sim- 
ple ratios. Thus, from the two simple ratios, 5 : 16 and 8 : 2, we 

5 : 16 
8 : 2 
may form the compound ratio -= — o~rTc — 9 or V 6 x § = f § = f • 

415. The Reciprocal of a ratio is 1 divided by the ratio ; or, 
which is the same thing, it is the antecedent divided by the con- 
sequent. Thus, the ratio of 7 to 9 is 7 : 9 or |, and its reciprocal 
is |. 

Note. — The quotient of the second term divided by the first is sometimes 
called a Direct Ratio, and the quotient of the first term divided by the second, 
an Inverse or Reciprocal Ratio, 

416. One quantity is said to vary directly as another, when 
the two increase or decrease together in the same ratio ; and one 
quantity is said to vary inversely as another, when one increases 
in the same ratio as the other decreases. Thus time varies directly 
as wages ; that is, the greater the time the greater the wages, and 
the less the time the less the wages. Again, velocity varies in- 
versely as the time, the distance being fixed; that is, in traveling 
a given distance, the greater the velocity the less the time, and the 
less the velocity the greater the time, 

417. Ratio can exist only between like numbers, or between 
two quantities of the same kind. But of two unlike numbers or 
quantities, one may vary either directly or inversely as the other. 
Thus, cost varies directly as quantity, in the purchase of goods; 
time varies inversely as velocity, in the descent of falling bodies. 
In all cases of this kind, the quantities, though unlike in kind, 
have a mutual dependence, or sustain to each other the relation 
of cause and effect. 



. 



RATIO. 245 

418. I n * ne comparison of like numbers we observe, 

I. If the numbers are simple, whether abstract or concrete, 
their ratio may be found directly by division. 

II. If the numbers are compound, they must first be reduced 
to the same unit or denomination. 

III. If the numbers are fractional, and have a common de- 
nominator, the fractions will be to each other as their numerators ; 
if they have not a common denominator, their ratio may be found 
either directly by division, or by reducing them to a common 
denominator and comparing their numerators. 

410. Since the antecedent is a divisor and the consequent a 
dividend, any change in either or both terms will be governed by 
the general principles of division, (117). We have only to sub- 
stitute the terms antecedent, consequent, and ratio, for divisor, 
dividend, and quotient, and these principles become 

GENERAL PRINCIPLES OF RATIO. 

Prin. I. Multiplying the consequent multiplies the ratio; divi- 
ding the consequent divides the ratio. 

Prin. II. Multiplying the antecedent divides the ratio ; dividing 
the antecedent multiplies the ratio. 

Prin. III. Multiplying or dividing both antecedent and conse- 
quent by the same number does not alter the ratio. 

420. These three principles may be embraced in one 

general law. 

A change in the consequent produces a LIKE change in the ratio ; 
htf a change in the antecedent produces an OPPOSITE change in the 
ratio. 

431* Since the ratio of two numbers is equal to the conse- 
quent divided by the antecedent, it follows, that 

I. The antecedent is equal to the consequent divided by the 
ratio; and that, 

II. The consequent is equal to the antecedent multiplied by the 
ratio. 

21* 



246 RATIO. 

EXAMPLES FOR PRACTICE. 

1. What part of 28 is 7? 

^ = J ; or, 28 : 7 as 1 : % ; that is, 28 has the same ratio to 7 that 

1 has to \. Ans. \. 

2. What part of 42 is 6? 

3. What is the ratio of 120 to 80 ? Ans. f . 

4. What is the ratio of 8| to 60 ? Ans. 7. 

5. What is the ratio of ^ to 26 ? 

6. What is the ratio of 7| to 2|? Ans. §°. 

7. What is the ratio of J to T % ? Ans. 41. 

8. What is the ratio of 1 mi. to 3 fur. ? Ans. f . 

9. What is the ratio of 1 wk. 3 da. 12 h. to 9 wk. ? Ans. 6. 

10. What is the ratio of 10 A. 1 K. 20 P. to 6 A. 2 R. 30 P. ? 

11. What is the ratio of 25 bu. 2 pk. 6 qt. to 40 bu. 4.5 pk. ? 

12. What is the ratio of 18f ° to 45' 30" ? 

12£ £ of 4 

13. What part of ~ is 5-^-i ? ^ws. ^. 

11^ 93 

14. What is the ratio of — f to f of T % of ^ ? .4?is. 5 <^. 

15. Find the reciprocal of the ratio of 42 to 28. Ans. 1|. 

16. Find the reciprocal of the ratio of 3 qt. to 43 gal. 

17. If the antecedent be 15 and the ratio |, what is the conse- 
quent? Ans. 12. 

18. If the consequent be 3| and the ratio 7, what is the ante- 
cedent? Ans. £|. 

19. If the antecedent be | of f and the consequent .75, what 
is the ratio ? 

20. If the consequent be $6.12| and the ratio 25, what is the 
antecedent? Ans. $.245. 

21. If the ratio be ^ and the antecedent |, what is the conse- 
quent ? 

22. If the antecedent be 13 A. 3 R. 25 P. and the ratio |f, 
what is the consequent ? Ans. 6 A. 2 R. 10 P. 



PROPERTIES OF PROPORTION. 247 



PROPORTION. 

422. Proportion is an equality of ratios. Thus, the ratios 
5 : 10 and 6 : 12, each being equal to 2, form a proportion. 

Note. — When four numbers form a proportion, they are said to be propor- 
tional. 

423. Proportion is indicated in three ways : 

1st. By a double colon placed between the two ratios; thus, 
3 : 4 : : 9 : 12 expresses the proportion between the numbers 3, 4, 
9, and 12, and is read, 3 is to 4 as 9 is to 12. 

2d. By the sign of equality placed between two ratios ; thus, 
3 : 4 = 9 : 12 expresses proportion, and may be read as above, or, 
the ratio of 3 to 4 equals the ratio of 9 to 12. 

3d. By employing the second method of indicating ratio; thus, 
| = ^ 2 indicates proportion, and may be read as either of the 
above forms. 

424. Since each ratio consists of two terms, every proportion 
must consist of at least four terms. Of these 

The Extremes are the first and fourth terms ; and 

The Means are the second and third terms. 

425* Three numbers are proportional when the first is to the 
second as the second is to the third. Thus, the numbers 4, 6, 
and 9 are proportional, since 4:6=6:9, the ratio of each couplet 
being |, or 1J. 

426* When three numbers are proportional, the second term 
is called the Mean Proportional between the other two. 

427. If we have any proportion, as 
3 : 15 = 4 : 20, 

Then, indicating this ratio by the second method, we have 

15 2 

% f • 

Reducing these fractions to a common denominator, 
15 X 4 _ 20 X 3 
12 12 " 

And since these two equal fractions have the same denominator, 
the numerator of the first, which is the product of the means, must 
be equal to the numerator of the second, which is the product of the 
extremes ; or, 15 X 4 = 20 X 3. Hence, 



248 



PROPORTION. 



I. In every proportion the product of the means equals the 
product of the extremes. 

igain, take any three terms in proportion, as 

4 : 6=6 : 9 
Then, since the product of the means equals the product of the ex- 
tremes, 

6 2 == 4 x 9. Hence, 

II. The square of a mean proportional is equal to the product 
of the other two terms. 

4L28. Since in every proportion the product of the means 
equals the product of the extremes, (4527, 1), it follows that, any 
three terms of a proportion being given, the fourth may be found 
by the following 

Rule. I. Divide the product of the extremes by one of the 
means, and the quotient will be the other mean. Or, 

II. Divide the product of the means by one of the extremes, and 
the quotient will be the other extreme. 

EXAMPLES FOR PRACTICE. 

The required term in an operation will be denoted by (?), 
which may be read " how many," or " how much/' 

Find the term not given in each of the following proportions : 

1. 4 : 26 =10 :(?). Ans. 65. 

2. $8865 : $720 = ( ? ) : 16 A. Ans. 197 A. 

3. Q yd. : ( ? ) : : $9.75 : $29.25. Ans. 13£ yd. 

4. (?) : 21 A. 3 R. 20 P. : : $1260 : $750. 

Ans. 36 A. 3 R. 

5. $7.50:18 = (?):7 T Voz. 

6. 7 oz :(?):: £30 : £407 2s. 10§d. Ans. 7 lb. 11 oz. 

7. ( ? ) : .15 hhd. : : $2.39 : $.3585. Ans. 1 hhd. 

8. 1 T. 7 cwt. 3 qr. 20 lb. : 13 T. 5 cwt. 2 qr. = $9.50 :.(?). 

9. $175.35 : (?) = | : f Ans. $601.20. 

10. (?) : $12£ 2404 : 149^/^. Ans. $20|. 

11. | yd. :(?):: *| : $59,062. Ans. 40£ yd. 



SIMPLE PROPORTION. 249 

CAUSE AND EFFECT. 

429. Every question in proportion may be considered as a 
comparison of two causes and two effects. Thus, if 3 dollars as 
a cause will buy 12 pounds as an effect, 6 dollars as a cause will 
buy 24 pounds as an effect. Or, if 5 horses as a cause consume 
10 tons as an effect, 15 horses as a cause will consume 30 tons as 
an effect. 

Causes and effects in proportion are of two kinds — simple and 
compound. 

430* A Simple Cause or Effect contains but one element ; 
as price, quantity, cost, time, distance, or any single factor used 
as a term in proportion. 

431. A Compound Cause or Effect is the product of two or 
more elements ; as the number of workmen taken in connection 
with the time employed, length taken in connection with breadth 
and depth, capital considered with reference to the time em- 
ployed, etc. 

433* Since like causes will always be connected with like 
effects, every question in proportion must give one of the following 
statements : 

1st Cause : 2d Cause = 1st Effect : 2d Effect. 
1st Effect : 2d Effect = 1st Cause : 2d Cause, 
in which the two causes or the two effects forming one couplet, 
must be like numbers and of the same denomination. 

Considering all the terms of a proportion as abstract numbers, 
we may say that 

1st Cause : 1st Effect = 2d Cause : 2d Effect. 

But as ratio is the result of comparing two numbers or things 
of the same hind, (417), the first form is regarded as the most 
natural and philosophical. 

SIMPLE PROPORTION. 

433. Simple Proportion is an equality of two simple ratios, 
and consists of four terms. 

Questions in simple proportion involve only simple causes and 
simple effects. 



250 



PROPORTION. 






FIRST METHOD. 

1. If $8 will buy 36 yards of velvet, how many yards may be 



bought for $12 ? 



STATEMENT. 

$ $ yds. 

8 : 12 = 36 ■ : 

1st cause. 2d cause. 1st effect. 
OPERATION. 

8 x (?) = 12 x 36 



yds. 

(?) 

2d effect. 



6 



(?) = 



>*X 



^9 



= 54 yd. 



Analysis. The re- 
quired term in this ex- 
ample is an effect ; and 
the statement is, $8 is 
to $12 as 36 yards is 
to ( ? ), or how many 
yards. Dividing 12 X 
36, the product of the 
means, by 8, the given 
extreme, we have ( ? ) 
= 54 yards, the re- 
quired term, (428, II). 
2. If 6 horses will draw 10 tons, how many horses will draw 
15 tons? 

Analysis. In this ex- 
ample a cause is required ; 
and the statement is, 6 
horses is to ( ? ), or how 
many horses, as 10 tons is 
to 15 tons. Dividing 15 X 
6, the product of the ex- 
tremes, by 10, the given 
mean, we have 9, the re- 
quired term, (428, I). 





STATEMENT. 




horses. 


horses. tons. 


tons. 


6 : 


(?) = 10 


: 15 


1st cause. 


2d cause. 1st effect. 


2d effect. 



OPERATION. 

(?) f 



() = 9 horses. 

434. Hence the following 

Rule. I. Arrange the terms in the statement so that the causes 
shall compose one couplet, and the effects the other, putting (?) in 
the place of the required term. 

II. If the required term be an extreme, divide the product of the 

means by the given extreme; if the required term be a mean, 

divide the product of the extremes by the given mean. 

Notes. — 1. If the terms of any couplet be of different denominations, they 
must be reduced to the same unit value. 

2. If the odd term be a compound number, it must be reduced either to its 
lowest unit, or to a fraction or a decimal of its highest unit. 

3. If the divisor and dividend contain one or more factors common to both, 
they should be canceled. If any of the terms of a proportion contain mixed 



SIMPLEJPROPORTION. 251 

numbers, they should first bo changed to improper fractions, or the fractional 
part to a decimal. 

4. When the vertical line is used, the divisor and (?) are written on the left, 
and the factors of the dividend on the right. 



SECOND METHOD. 

435. The following method of solving examples in simple 
proportion without making the statement in form, may be used 
by those who prefer it. 

Every question in simple proportion gives three terms to find a 
fourth. Of the three given terms, two will always be like numbers, 
forming the complete ratio, and the third will be of the same name or 
kind as the required term, and may be regarded as the antecedent of 
the incomplete ratio ; hence the required term may be found by mul- 
tiplying this third term, or antecedent, by the ratio of the other two, 
(428, II). 

Prom the conditions of the question we can readily determine 
whether the answer, or required term, will be greater or less than the 
third term ; if greater, then the ratio will be greater than 1, and the 
two like numbers must be arranged in the form of an improper frac- 
tion, as a multiplier ; if less, then the ratio will be less than 1, and 
the two like numbers must be arranged in the form of a proper frac- 
tion, as a multiplier. 

1. If 4 tons of hay cost $36, what will 5 tons cost? 

operation. Analysis. In this example, 4 

$36 X - = $45 Arts. tons an d 5 tons are the like terms, 

and $36 is the third term, and of 
the same kind as the answer sought. Now if 4 tons cost $36, will 5 
tons cost more, or less, than $36 ? Evidently more : and the required 
term will be greater than the third term, $36, and the ratio greater 
than 1. We therefore arrange the like terms in the form of an im- 
proper fraction, |, for a multiplier, and obtain $45, the answer. 
- 2. If 7 men build 21 rods of wall in a day, how many rods will 
4 men build in the same time ? 

operation. Analysis. In this example, 7 

2\ x 4 __. \2 rods Arts. men an< ^ ^ men are * ne ^ e terms, 

and 21 rods is the third term, and 
of the same kind as the answer sought. Since 4 men will perform less 
work than 7 men in the same time, the required term will be less than 



252 PROPORTION. 

21, and the ratio less than 1. We therefore arrange the like terms 
in the form of a proper fraction, 4, and obtain by multiplication, 12 
rods, the answer. 

430. Hence the following 

Rule. I. With the two given numbers, which are of the same 
name or kind, form a ratio greater or less than 1, according as the 
answer is to be greater or less than the third given number. 

II. Multiply the third number by this ratio ; the product will be 
the required number or answer. 

Notes. — 1. Mixed numbers should first be reduced to improper fractions, and 
the ratio of the fractions found according to 418. 
2. Reductions and cancellation may be applied as in the first method. 

The following examples may be solved by either of the fore- 
going methods. 

EXAMPLES FOR PRACTICE. 

1. If 12 gallons of wine cost $30, what will 63 gallons cost? 

2. If 9 bushels of wheat make 2 barrels of flour, how many 
barrels of flour will 100 bushels make ? Ans. 450. 

3. If 18 bushels of wheat be bought for $22.25, and sold for 
$26.75, how much will be gained on 240 bushels, at the same rate 
of profit? Ans. $60. 

4. If 6 J bushels of oats cost $3, what will 9} bushels cost? 

5. What will 87.5 yards of cloth cost, if If yards cost $.42 ? 

6. If by selling $1500 worth of dry goods I gain $275.40, what 
amount must I sell to gain $1000 ? 

7. If 20 men can perform a piece of work in 15 days, how 
many men must be added to the number, that the work may be 
accomplished in | of the time ? Ans. 5. 

8. If 100 yd. of broadcloth cost $473.07 T 9 3, how much will 
3.25 yd. cost? 

9. If 1 lb. 4 oz. 10 pwt. of gold may be bought for $260.70, 
how much may be bought for $39.50 ? Ans. 2 oz. 10 pwt. 

10. In what time can a man pump 54 barrels of water, if he 
pump 24 barrels in 1 h. 14 min. ? Ans. 2 h. 46 min. 30 sec. 

11. If | of a bushel of peaches cost $£§, what part of a bushel 
can be bought for fy^ ? Ans. T 7 2 bu. 



COMPOUND PROPORTION. 253 

12. If the annual rent of 46 A. 3 R. 14 P. of land be 8374.70, 
how much will be the rent of 35 A. 2 R. 10 P. ? 

13. If a man gain $1870.65 by his business in 1 yr. 3 mo., how 
much would he gain in 2 yr. 8 mo., at the same rate ? 

14. Two numbers are to each other as 5 to 7 J, and the less is 
164.5, what is the greater ? Atis. 246.75. 

15. If 16 head of cattle require 12 A. 3 R. 36 P. of pasture 
during the season, how many acres will 132 head of cattle require ? 

An*. 107 A. 7 P. 

16. If a speculator in grain gain $26.32 by investing $325, how 
much would he gain by investing $2275.50 ? 

17. What will be the cost of paving an open court 60.5 ft. long 
and 44 ft. wide, if 14.25 sq. yd. cost $341? 

18. At 6J cents per dozen, what will be the cost of lOf gross 
of steel pens ? 

19. If when wheat is 7s. 6d. per bushel, the bakers' loaf will 
weigh 9 oz., what ought it to weigh when wheat is 6s. per bushel ? 

Ans. Hi oz. 



COMPOUND PROPORTION. 

437* Compound Proportion is an expression of equality be- 
tween a compound and a simple ratio, or between two compound 
ratios. 

It embraces the class of questions in which the causes, or the 
effects, or both, are compound. The required term must be either 
a simple cause or effect, or a single element of a compound cause 
or effect. 

FIRST METHOD. 

1. If 8 men mow 40 acres of grass in 3 days, how many acres 
will 9 men mow in 4 days ? 

STATEMENT. 
1st cause. 2d cause. 1st effect. 2d effect. 

{3= {4 = 40 : (?) 

Or, 8 x 3:9 x 4 = 40 : (?) 

22 



254 PROPORTION. 

operation. Analysis. In this ex- 

9x4x40 /»a j ample the required term 

' * ^ 8~X~3 = ' is the second effect ; and the 

statement is, 8 men 3 days 
is to 9 men 4 days, as 40 acres is to ( ? ), or how many acres. Dividing 
the continued product of all the elements of the means by the ele- 
ments of the given extreme, we obtain ( ? ) = 60 acres. 

2. If 6 compositors in 14 hours can set 36 pages of 56 lines 
each, how many compositors, in 12 hours, can set 48 pages of 54 
lines each ? 

STATEMENT. 



1st cause. 2d cause. 1st effect. 2d effect. 

6 J (?) .. J 36 | 48 



f 6 . f (?) f 36 . f 
\ 14 ' I 12 ' ' } 56 * I 



operation. Analysis. In this example, an element of 

the second cause is required ; and the state- 
ment is, 6 compositors 14 hours is to ( ?) com- 
positors 12 hours as 36 pages of 56 lines each 
is to 48 pages of 54 lines each. Now, since the 



(?) 



/ \ _ 9 j LflSt required term is an element of one of the means, 
we divide the continued product of all the ele- 
ments of the extremes by the continued product of all the given ele- 
ments of the means. Placing the dividend on the right of the verti- 
cal line and the divisors on the left, and canceling equal factors we 
obtain ( ? ) = 9. 

438; From these illustrations we deduce the following 

Rule. I. Of the given terms, select those which constitute the 

causes, and those which constitute the effects, and arrange them in 

couplets, putting (?) in place of the required term. 

II. Then, if the blank term (?) occur in either of the extremes, 

divide the product of the means by the product of the extremes; 

but if the blank term occur in either mean, divide the product of 

the extremes by the product of the means. 

Notes. — 1. The causes must be exactly alike in the number and hind of their 
terms ; the same is true of the effects. 

2. The same preparation of the terms by reduction is to be observed as in 
simple proportion. 



COMPOUND PROPORTION. 255 



SECOND METHOD. 



439. The second method given in Simple Proportion, is also 
applicable in Compound Proportion. 

In every example in compound proportion all the terms appear in 
couplets, except one, called the odd term, which is always of the same 
kind as the answer sought. Hence the required term in a compound 
proportion may be found, by multiplying the odd term by the com- 
pound ratio composed of all the simple ratios formed by these couplets, 
each couplet being arranged in the form of a fraction. 

The fraction formed by any couplet will be improper when the re- 
quired term, considered as depending on this couplet alone, should 
be greater than the odd term ; and proper, when the required term 
should be less than the odd term. 

1. If it cost $4320 to supply a garrison of 32 men with pro- 
visions for 18 days, when the rations are 15 ounces per day, what 
will it cost to supply a garrison of 24 men 34 days, when the 
rations are 12 ounces per day ? 

OPERATION. 

men. days. ounces. 

$4320 X M X f| X j| = $4896 

by cancellation. Analysis. In this example there are 

(?) 4320 three pairs of terms, or couplets, viz., 32 

32 24 men and 24 men, 18 days and 34 days, 15 

18 34 ounces and 12 ounces ; and there is an odd 

15 12 term, $4320, which is of the same kind as 

( ) = $4895 Arts, the required term. We arrange each coup- 
let as a multiplier of this term, thus; 
Tirst, if it cost $4320 to supply 32 men, will it cost more, or less, to 
supply 24 men ? Less ; we therefore arrange the couplet in the form of 
a proper fraction as a multiplier, and we have $4320 X ff . Next, if it 
cost $4320 to supply a garrison 18 days, will it cost more, or less, to 
supply it 34 days ? More ; hence the multiplier is the improper frac- 
tion ?4, and we have $4320 X J| X ft. Next, if it cost $4320 to 
supply a garrison with rations of 15 ounces, will it cost more, or less, 
when the rations are 12 ounces? Less; consequently, the multiplier 
is the proper fraction j§, and we have $4230 X || X ft X f| =$4896, 
the required term. Hence the following 



256 PROPORTION. 

Rule. I. Of the terms composing each couplet form a ratio 
greater or less than 1, in the same manner as if the answer de- 
pended on those two and the third or odd term. 

II. Multiply together the third or odd term and these ratios; 
the product will be the answer sought 

EXAMPLES FOR PRACTICE. 

1. If 12 horses plow 11 acres in 5 days, how many horses 
would plow 33 acres in 18 days ? Ans. 10. 

2. If 480 bushels of oats will last 24 horses 40 days, how long 
will 300 bushels last 48 horses, at the same rate ? 

Ans. 12} days. 

3. If 7 reaping machines can cut 1260 acres in 12 days, in 
how many days can 16 machines reap 4728 acres ? 

Ans. 19.7 days. 

4. If 144 men in 6 days of 12 hours each, build a wall 200 ft. 
long, 3 ft. high, and 2 ft. thick, in how many days of 7 hours 
each can 30 men build a wall 350 ft. long, 6 ft. high, and 3 ft. 
thick? Ans. 7.2 da. 

5. In how many days will 6 persons consume 5 bu. of potatoes, 
if 3 bu. 3 pk. last 9 persons 22 days? 

6. How many planks lOf ft. long and 1} in. thick, are equiva- 
lent to 3000 planks 12 ft. 8 in. long and 2| in. thick ? 

Ans. 6531 h 

7. If 300 bushels of wheat @ $1.25 will discharge a certain 
debt, how many bushels @ $.90 will discharge a debt 3 times as 
great ? Ans. 1250 bu. 

8. If 468 bricks, 8 inches long and 4 inches wide, are required 
for a walk 26 ft. long and 4 ft. wide, how many bricks will be 
required for a walk 120 ft. long and 6 ft. wide ? 

9. If a cistern 17} ft. long, 10} ft. wide, and 13 ft. deep, hold 
546 barrels, how many barrels will a cistern hold that is 16 ft. 
long, 7 ft. wide, and 15 ft. deep ? Ans. 384 bbl. 

10. If 11 men can cut 147 cords of wood in 7 days, when they 
work 14 hours per day, how many days will it take 5 men to cut 
150 cords, working 10 hours each day ? 



MISCELLANEOUS EXAMPLES. 257 

PROMISCUOUS EXAMPLES IN PROPORTION. 

1. If a staff 4 ft. long cast a shadow 7 ft. in length, what is 
the hight of a tower that casts a shadow of 198 ft. at the same 
time ? Ans. 113 * ft. 

2. A person failing in business owes $972, and his entire prop- 
erty is worth but $607.50; how much will a creditor receive on a 
debt of $11.33i? Arts. $7.08+. 

3. If 3 cwt. can be carried 660 mi. for $4, how many cwt. can 
be carried 60 mi. for $12 ? Ans. 99. 

4. A man can perform a certain piece of work in 18 days by 
working 8 hours a day; in how many days can he do the same 
work by working 10 hours a day ? Ans. 14|. 

5. How much land worth $16.50 an acre, should be given in 
exchange for 140 acres, worth $24.75 an acre? 

6. If I gain $155.52 on $1728 in 1 yr. 6 mo., how much will 
I gain on $750 in 4 yr. 6 mo.? Ans. $202.50. 

7. If 1 lb. 12 oz. of wool make 2 i yd. of cloth 6 qr. wide, how 
many lb. of wool will it take for 150 yd. of cloth 4 qr. wide ? 

8. What number of men must be employed to finish a piece of 
work in 5 days, which 15 men could do in 20 days? Ans. 60. 

9. At 12s. 7d. per oz., N. Y. currency, what will be the cost 
of a service of silver plate weighing 15 lb. 11 oz. 13 pwt. 17 gr. ? 

10. If a cistern 16 ft. long, 7 ft. wide, and 15 ft. deep, cost 
$36.72, how much, at the same rate per cubic foot, would another 
cistern cost that is 17i ft. long, 10 J ft. wide, and 16 ft. deep? 

11. A borrows $1200 and keeps it 2 yr. 5 mo. 5 da. ; what 
sum should he lend for 1 yr. 8 mo. to balance the favor ? 

12. A farmer has hay worth $9 a ton, and a merchant has flour 
worth $5 per barrel. If in trading the former asks $10.50 for 
his hay, how much should the merchant ask for his flour ? 

13. If 12 men, working 9 hours a day for 15| days, were able 
to execute f of a job, how many men may be withdrawn and the 
job be finished in 15 days more, if the laborers are employed 
only 7 hours a day ? Ans. 4. 

14. If the use of $300 for 1 yr. 8 mo. is worth $30, how much 
is the use of $210.25 for 3 yr. 4 mo. 24 da. worth ? 

22* r 



258 PROPORTION. 

15. What quantity of lining | yd. wide, will it require to line 
9} yd. of cloth, 1J yd. wide? Ans. 15| yd. 

16. If it cost $95.60 to carpet a room 24 ft. by 18 ft., how 
much will it cost to carpet a room 38 ft. by 22 ft. with the same 
material? Ans. $185.00+. 

17. If 16 T 7 5 cords of wood last as long as ll 2 9 g tons of coal, 
how many cords of wood will last as long as 15 T ^ tons of coal ? 

18. A miller has a bin 8 ft. long, 4i ft. wide, and 2| ft. deep, 
and its capacity is 75 bu. ; how deep must he make another bin 
which is to be 18 ft. long and 3| feet wide, that its capacity may 
be 450 bu. ? Ans. 7^ ft. 

19. If 4 men in 2} days, mow 6| acres of grass, by working 
8J hours a day, how many acres will 15 men mow in 3f days, by 
working 9 hours a day ? Ans. 40{£ acres. 

20. If an army of 600 men have provisions for 5 weeks, allowing 
each man 12 oz. a day, how many men may be maintained 10 
weeks with the same provisions, allowing each man 8 oz. a day ? 

21. A cistern holding 20 barrels has two pipes, by one of which 
it receives 120 gallons in an hour, and by the other discharges 
80 gallons in the same time ; in how many hours will it be filled ? 

22. A merchant in selling groceries sells 14 T 9 g oz. for a pound; 
how much does he cheat a customer who buys of him to the amount 
of $38.40 ? Ans. $3.45. 

23. If 5 lb. of sugar costs $.62}, and 8 lb. of sugar are worth 
5 lb. of coffee, now much will 75 lb. of coffee cost? 

24. B and C have each a farm ; B's farm is worth $32.50 an 
acre, and C's $28.75 ; but in trading B values his at $40 an acre. 
What value should C put upon his ? 

25. If it require 4950 reams of paper to print 12000 copies of 
an 8vo. book containing 550 pages, how many reams will be required 
to print 3000 copies of a 12mo. book containing 320 pages ? 

26. If 248 men, in 5} days of 12 hours each, dig a ditch of 7 
degrees of hardness, 232} yd. long, 3# yd. wide, and 2 J yd. deep; 
in how many days of 9 hours each, will 24 men dig a ditch of 4 
degrees of hardness, 387} yd. long, 5^ yd. wide, and 3} yd, deep? 

Ans. 155. 



NOTATION. 259 



PERCENTAGE. 

4:4:0. Per Cent, is a contraction of the Latin phrase per 
centum, and signifies by the hundred; that is, a certain part of 
every hundred, of any denomination whatever. Thus, 4 per cent 
means 4 of every hundred, and may signify 4 cents of every 100 
cents, 4 dollars of every 100 dollars, 4 pounds of every 100 
pounds, etc. 

NOTATION. 

441. The character, %, is generally employed in business 
transactions to represent the words per cent ; thus 6 °fc signifies 
6 per cent. 

442. Since any per cent, is some number of hundredths, it is 
properly expressed by a decimal fraction; thus 5 per cent. 
= 5 °fo = .05. Per cent, may always be expressed, however, 
either by a decimal or a common fraction, as shown in the following 

TAJBLE. 

Words. Symbols. Decimals. Common fractions. 

1 per cent. = 1 % = .01 = T ^ = yfo 

2 per cent. = 2 % = .02 = T f ^ = fa 

4 percent. = 4 % = .04 = T J T = 'fa < 

5 per cent. = 5 % = .05 = T ^ = fa 

6 percent. = 6 % = .06 = T fo = A 

7 per cent. = 7 % = .07 == yj^ = jfo 

8 percent. = 8 % = .08 = yfo = al 
10 percent. = 10 <f = .10 = ^ = T V 
20 percent. = 20 % = .20 = ^ = i 
25 per cent. = 25 % = .25 = T 2 <& = \ 
50 percent. = 50 % = .50 = T %% = J 

100 percent. = 100 % = 1.00 = T ini = 1 

125 percent. = 125 % = 1.25 ** f g§ = | 

J per cent. = J # = .005 = r /^ = ^ 

| per cent. = f % = .0075 jjfa = T 5 

12} per cent. = 12} % = .125 = fa'fa = i 



260 PERCENTAGE. 

EXAMPLES FOR PRACTICE. 

1. Express decimally 3 per cent. ; 9 per cent. ; 12 per cent. ; 
16 per cent. ; 23 per cent. ; 37 per cent. ; 75 per cent. ; 125 per 
cent. ; 184 per cent. ; 205 per cent. 

2. Express decimally 15 % ; 11 % j 4J % j 5* % } 8f % ; 
20* % j 25f % ; 35| cf } 24* % ; 130* %. 

3. Express decimally i per cent. ; f per cent. ; \ per cent. ; 
| per cent. ; | per cent. ; jfe per cent. ; T 3 ^ per cent. ; 1 T 5 ^ per 
cent. ] 104 per cent. 

4. Express by common fractions, in their lowest terms, 4 % ; 
37* %; 16* %; 114%; 42f %; 45 T \ %; 43^ %. 

5. What per cent, is .0725 ? 

Analysis. .0725 = .07i *= 7£ Jg, ^Ircs. 

6. What per cent, is .065? ^4ws. 6* %. 

7. What per cent, is .14375? Arts. 14| %. 

8. What per cent, is .0975 ? 

9. What per cent, is .014 ? 

10. What per cent, is .1025? 

11. What per cent, is .004 ? 

12. What per cent, is .028 ? 

13. What % is .1324? 

14. What % is .084* ? 

15. What % is .004 T <y? Ans. T \ % 

16. What % is .003-^ ? 



GENERAL PROBLEMS IN PERCENTAGE. 

If 51. In the operations of Percentage there are five parts or 
elements, namely : Rate per cent., Percentage, Base, Amount, and 
Difference. 

444. Rate per Cent., or Rate, is the decimal which denotes 
how many hundredths of a number are to be taken. 

Notes. — 1. Such expressions as 6 per cent., and 5 % , are essentially decimals, 
the words per cent., or the character %, indicating the decimal denominator. 

2. If the decimal be reduced to a common fraction in its lowest terms, this 
fraction will still be the equivalent rate, though not the rate per cent. 



PROBLEMS IN PERCENTAGE. 



261 



44o« Percentage is that part of any number which is indi- 
cated by the rate. 

446. The Base is the number on which the percentage is 
computed. 

447. The Amount is the sum obtained by adding the per- 
centage to the base. 

448. The Difference is the remainder obtained by subtract- 
ing the percentage from the base. 

PROBLEM I. 

449. Given, the base and rate to find the per- 
centage. 



1. What is 5 <f of 360? 



OPERATION. 

360 
.05 



18.00, Ans. 

Or, 

360 x 2\> = 18, Arts. 



Analysis. Since 5 fo of any 
number is .05 of that number, 
(442), we multiply the base, 360, 
by the rate, .05, and obtain the 
percentage, 18. Or, since the rate 
is y^ff = -^, we have 360 X / ¥ = 
18, the percentage. Hence the fol- 
lowing 



RULE. Multiply the base by the rate. 

Note 1. — Percentage is always a product, of which the base and rate are the 
factors. * 



1. What 

2 What 

3. What 

4. What 

5. What 

6. What 

7. What 

8. What 

9. What 
10. What 



EXAMPLES FOR PRACTICE. 

s 4 per cent, of 250 ? 
s 7 per cent, of 3500 ? 
s 16 per cent, of 324 ? 
s 12! per cent, of $5600? 
s9 % of 785 lbs.? 
s25 % of 960 mi.? 
s75 % of 487 bu.? 
s33| % of 2757 men? 
sl25 % of 756? 
si % of $2364? 



Ans. 10. 

Ans. 245. 

Ans. 51.84. 

Ans. 70. 



Ans. 365.25 bu. 



Ans. 5.91. 



262 PERCENTAGE. 

11. What is 3| c/o of $856? Ans. $31.38 +. 

12. What is | % of |? Ans. T fa. 

13. What is 14f % of 5| ? 

14. If the base is $375, and the rate .05, what is the percent- 
age? Ans. $18.75. 

15. A man owed $536 to A, $450 to B, and $784 to C; how 
much money will be required to pay 54 % of his debts ? 

16. My salary is $1500 a year; if I pay 15 % for board, 5 % 

for clothing, 6 °f for books, and 8 °Jo f° r incidentals, what are 

my yearly expenses ? Ans, $510. 

Note 2. — 15 % + 5 % + 6 % + 8 % = 34 % . In all cases where several 
rates refer to the same base, they may be added or subtracted, according to the 
conditions of the question. 

17. A man having a yearly income of $3500, spends 10 per 
cent of it the first year, 12 per cent, the second year, and 18 per 
cent, the third year ; how much does he save in the 3 years ? 

18. A had $6000 in a bank. He drew out 25 % of it, then 
30 °J C of the remainder, and afterward deposited 10 % of what 
he had drawn ; how much had he then in bank ? Ans. $3435. 

19. A merchant commenced business, Jan. 1, with a capital of 
$5400, and at the end of 1 year his ledger showed the condition 
of his business as follows: For Jan., 2 % gain; Feb., *&\ °/o g a in; 
March, \ % loss; Apr., 2 % gain; May, 2 J (f gain; June, If 
cfc loss ; July, 1} % gain ; Aug., 1 % loss ; Sept., 2f % gain ; 
Oct., 4 % gain; Nov., f % loss; Dec, 3 <f gain. What were 
the net profits of his business for the year? Ans. $918. 

PROBLEM II. 

450. Given, the percentage and base to find the 
rate, 

1. What per cent of 360 is 18 ? 

operation. Analysis. Since the percent- 

18 -i- 360 = .05 = 5 °f age is always the product of the 

Q r base and rate, (449), we divide 

i 8_ _ ) __ 05 = 5 of ^ ie gi yen Percentage, 18, by the 

3 6 0—20 ' /O given bag ^ 360 ^ and obtain ^ 

required rate, .05 = 5 % . Hence the 



PROBLEMS IN PERCENTAGE. 263 

Rule. Divide the percentage by the base. 

EXAMPLES FOR PRACTICE. 

1. What per cent, of $720 is $21.60 ? Ans. 3. 

2. What per cent, of 1560 lb. is 234 lb. ? 

3. What per cent, of 980 rd. is 49 rd. ? 

4. What per cent, of £320 10s. is £18. 8.8s. ? Ans. 8. 

5. What per cent, of 46 gal. is 5 gal. 3 qt. ? Ans. 12}. 

6. What per cent, of 7.85 mi. is 5.495 mi.? Ans. 70, 

7. What per cent, of T 8 ¥ is § ? Ans. 75. 

8. What per cent, of | is ^ ? 

9. What per cent, of 560 is 80? 

10. The base is $578, and the percentage is $26.01 ; what is the 
rate? Ans. 4} f . 

11. The base is $972.24, and the percentage is $145,836; what 
is the rate ? 

12. An editor having 5600 subscribers, lost 448 ; what was his 
loss per cent? Ans. 8. 

13. A merchant owes $7560, and his assets are $4914 ; what 
per cent, of his debts can he pay ? Ans. 65. 

14. A man shipped 2600 bushels of grain from Chicago, and 
455 bushels were thrown overboard during a gale ; what was the 
rate per cent, of his loss ? 

15. A miller having 720 barrels of flour, sold 288 barrels; what 
per cent, of his stock remained unsold ? Ans. 60. 

16. What per cent, of a number is 30 % of | of it ? 

17. The total expenditures of the General Government, for the 
year ending June 30, 1858, were $83,751,511.57; the expenses 
of the War Department were $23,243,822.38, and of the Navy 
Department, $14,712,610.21. What per cent, of the whole ex- 
pense of government went for armed protection ? 

Ans. 4 5 J, nearly. 

18. In the examination of a class, 165 questions were sub- 
mitted to each of the 5 members ; A answered 130 of them, B 
125, C 96, D 110, and E 160. What was the standing of the 
class? Ans. 75.2 %. 



264 PERCENTAGE. 

PROBLEM III. 

451. Given, the percentage and rate, to find the 
base. 

1. 18 is 5 % of what number? 

AT ,„ nimTA „ Analysis. Since the percent- 

OPERATION. , T a 

age is always the product of the 
18 -=- .05 = 360, Ans. bage and ^ (440)> we diyide 

Or, the given percentage, 18, by the 

18 -^- ^ = 360, Ans. given rate, .05, or -fa, and obtain 

the base, 360. Hence the 
Rule. Divide the percentage hy the rate. 



EXAMPLES FOR PRACTICE. 

1. 18 is 25 % of what number? Ans. 72. 

2. 54 is 15 % of what number ? 

3. 17.5 is 2i cfo of what number? Ans. 750. 

4. 2.28 is 5 % of what number? 

5. 414 is 120 % of what number? 

6 6119 is 105} % of what number? Ans. 5800. 

7. .41 is 71f % of what number? Ans. .6. 

8. The percentage is $18.75, and the rate is 2} % ) what is the 
base? Ans. $750. 

9. The percentage is 311, and the rate 31} % ; w ^ at * s the 
base? 

10. I sold my house for $4578, which was 84 % of its cost; 
what was the cost ? Ans. $5450. 

11. A wool grower sold 3150 head of sheep, and had 30 % of 
his original flock left ; how many sheep had he at first ? 

12. A man drew 40 % of his bank deposits, and expended 13 J 
cj of the money thus drawn in the purchase of a carriage worth 
$116; how much money had he in bank? Ans. $2175. 

13. If $147.56 is 13-J % of A's money, and 4| % of A's 
money is 8 <f of B's, how much more money has A than B ? 

Ans. $461.12}. 



PROBLEMS IN PERCENTAGE. 265 

14. In a battle 4 % of the army were slain upon the field ; and 5 
<f of the remainder died of wounds, in the hospital. The differ- 
ence between the killed and the mortally wounded was 168 ; how 
many men were there in the army ? Arts. 21000. 

Note.— 100 % — 4 % = 96 %, left after the battle; and 5 % of 96 % = 
44 % , the part of the army that died of wounds. 

15. A owns f of a prize and B the remainder; after A has 
taken 40 % of his share, and B 20 % of his share, the remainder 
is equitably divided between them by giving A $1950 more than 
B ; what is the value of the prize ? Arts. $7800. 



PROBLEM IV. 

452. Given, the amount and rate, to find the base. 

1. What number increased by 5 % of itself is equal to 378 ? 

operation. Analysis. If any number 

j j_ .05 = 1.05 ^e i ncre ased by 5 % of itself 

378 -i- 1.05 = 360, Ans. the amount will be 1.05 times 

the number. We therefore di- 
Or, vide the given amount, 378, by 

1 -L JL — |i 1.05, or f A, and obtain the base, 

378 -;- 1 1 = 360, Ans. 360, which is the number re- 

quired. Hence the 

Rule. Divide the amount by 1 plus the rate. 

Note 1. — The amount is always a product, of which the base is one factor, 
and 1 plus the rate the other factor. 

EXAMPLES FOR PRACTICE. 

1. What number increased by 15 % of itself is equal to 644? 

Ans. 560. 

2. A has §815.36, which is 4 % more than B has; how much 
money has B ? Ans. $784. 

3. Having increased my stock in trade by 12 % of itself, I 
find that I have $3800 ; how much had I at first ? 

4. In 1860 the population of a certain city was 39600, which 
was an increase of 10 % during the 10 years preceding; what 
was the population in 1850 ? 

23 



266 



PERCENTAGE. 



5. My crop of wheat this year is 8 % greater than my crop of 
last year, and I have raised during the two years 5200 bushels; 
what was my last year's crop? A?is. 2500 bu. 

Note 2. — 1.00 + 1.08 = 2.08. Hence, 5200 bu. = 2.08 % of last year's crop. 

6. The net profits of a nursery in two years were $6970, and 
the profits the second year were 5 % greater than the profits the 
first year ; what were the profits each year ? 

Ans. 1st year, $3400; 2d year, $3570. 

7. If a number be increased 8 %, and the amount be increased 

7 °Joi tne result will be 86.67; required the number. 

Note 3. — The whole amount will be 1.08 X 1.07 = 1.1556 times the original 
number. 

8. A produce dealer bought grain by measure, and sold it by 
weight, thereby gaining 1J <f in the number of bushels. He 
sold at a price 5 % above his buying price, and received $4910.976 
for the grain; required the cost. Ans. $4608. 

9. B has 6 %, and C 4 % more money than A, and they all 
have $11160 ; how much money has A ? Ans. $3600. 

10. In the erection of a house I paid twice as much for mate- 
rial as for labor. Had I paid 6 % more for material, and 9 % more 
for labor, my house would have cost $1284 ; what was its cost ? 

Ans. $1200. 

PROBLEM V. 

453. Given, the difference and rate to find the base. 

1. What number diminished by 5 % of itself, is equal to 342 ? 

operation. Analysis. If any number be di- 

1 .05 = .95 minished by 5 % of itself, the dif- 

342 ~- .95 = 360, Ans. ference will be .95 of the number. 

Or We therefore divide the given differ- 

1 — ^ s ™ ence, 342, by .95, or ff , and obtain 

342 -i- * 9 = 360, Ans. the base, 360, which is the required 

number. Hence the 

Rule. Divide the difference hy 1 minus the rate. 

Note. — The difference is always a product, of which the base is one factor, 
and 1 minus the rate the other. 



PROBLEMS IN PERCENTAGE. Ofjf 

EXAMPLES FOR PRACTICE. 

1. What number diminished by 10 % of itself is equal to 504? 

Arts. 560. 

2. The rate is 8 %, and the difference §4.87; what is the 
base ? 

3. After taking away 15 % of a heap of grain, there remained 
40 bu. 3^ pk. ; how many bushels were there at first ? 

Arts. 48 bu. 

4. Having sold 36 <f of my land, I have 224 acres left; how 
much land had I at first ? 

5. After paying 65 % of my debts, I find that §2590 will dis- 
charge the remainder ; how much did I owe in all ? 

Arts. §7400. 

6. A young man having received a fortune, deposited 80 % of 
it in a bank. He afterward drew 20 % of his deposit, and then 
had §5760 in bank ; what was his entire fortune ? 

Ans. §9000. 

7. A man owning i of a ship, sold 12 % of his share to A, and 
the remainder to B, at the same rate, for §20020 ; what was the 
estimated value of the whole ship ? Ans. §26000. 

8. An army which has been twice decimated in battle, now 
contains only 6480 men; what was the original number in the 
army? ' Ans. 8000. 

9. Each of two men, A and B, desired to sell his horse to C. 
A asked a certain price, and B asked 50 % niore. A then re- 
duced his price 20 %, and B his price 30 %, at which prices C 
took both horses, paying for them §148 ; what was each man's 
asking price ? . f A, §80. 

m ' IB, §120. 

10. A buyer expended equal sums of money in the purchase of 
wheat, corn, and oats. In the sales, he cleared 6 % on the wheat, 
and 3 % on the corn, but lost 17 % on the oats; the whole 
amount received was §2336. What sum did he lay out in each 
kind of grain? Ans. §800. 



268 PERCENTAGE. 

APPLICATIONS OF PERCENTAGE. 

454. The principal applications of Percentage, where time is 
not considered, are Commission, Stocks, Profit and Loss, Insurance, 
Taxes, and Duties. And since the five problems in Percentage 
involve all the essential relations of the parts or elements, we have 
for the above applications the following 

General Rule. Note what elements of Percentage are given 
in the example, and what element is required ; then apply the spe- 
cial rule for the corresponding case, 

COMMISSION. 

455. An Agent, Factor, or Broker, is a person who trans- 
acts business for another. 

45©. A Commission Merchant is an agent who buys and 
sells goods for another. 

457. Commission is the fee or compensation of an agent, 
factor, or commission merchant. 

458. A Consignment is a quantity of goods sent to one person 
to be sold on commission for another person. 

459. A Consignee is a person who receives goods to sell for 
another; and 

4©©. A Consignor is a person who sends goods to another to 
be sold. 

461. The Net Proceeds of a sale or collection is the sum 

left, after deducting the commission and other charges. 

Notr. — A person who is employed in establishing mercantile relations between 
others living at a distance from each other, is called the Correspondent of the 
party in whose behalf he acts. A correspondent is the agent of those whose 
custom or patronage he secures to the party in whose interest he is employed. 

463. Commission is usually reckoned at a certain per cent, of 
the money involved in the transaction ; hence we have the follow- 
ing relations : 

T. Commission is percentage, (445). 

II. The sum received by the agent as the price of property sold, 
or the sum invested by the agent in the purchase or exchange of 
property, is the base of commission, (446). 



COMMISSION. 269 

III. The sum remitted to an agent, and including both the pur- 
chase money and the agent's commission, is the amount, (447). 

IV. The sum due the employer or consignor as the net proceeds 
of a sale or collection, is the difference, (448). 

EXAMPLES FOR PRACTICE. 

1. My agent sells goods to the amount of $6250; what is his 
commission at 3 % ? 

operation. Analysis. According to 

$6250 X .03 == $187.50 Prob.J, (449), we multi- 

ply the sum obtained for 
the goods, $6250, which is the base of the commission, (II), by the 
rate of the commission, .03, and obtain the commission or percent- 
age, $187.50. 

2. A flour merchant remits to his agent in Chicago $3792, for 
the purchase of grain, after deducting the commission at 4 % ; 
how much will the agent expend for his employer, and what will 
be his commission ? 

operation. Analysis. Ac- 

1.00 + .04 = 1.04 cording to Prob. 

$3792 -T- 1.04 = $3650, for grain, IV, (452), we di- 

$3792 — $3650 = $142, commission. vide the remittance, 

$3792, which is 

amount, (III), by 1 plus the rate of commission, or 1.04, and obtain 

the base of commission, $3650, which is the sum to be expended in 

the purchase. Subtracting this from the remittance, we have $142, 

the commission. 

Note 1. — It is evident that the whole remittance, $3792, should not be taken 
as the base of commission; for that would be computing commission on commis- 
sion. A person must charge commission only on what he expends or collects, in 
his capacity as agent. 

3. A factor sold real estate on commission of 5 %, and returned 
to the owner, as the net proceeds, $8075; for what price did he 
sell the property, and what was his commission ? 

operation. Analysis. According to 

1.00 — .05 = .95 ProD - y > (453), we divide 

$8075 -f- .95 = $8500, price. the net proceeds, $8075, 

$8500 — $8075 = $425, com. which is difference, (IV), 

by 1 minus the rate of 
23* 



270 PERCENTAGE, 

commission, and obtain the base, $8500, which is the price of the 
property sold ; whence by subtraction, we obtain the commission, 
$425. 

4. An agent sold my house and lot for $8600 ; what was his 
commission at 2 J- % ? Ans. $193.50. 

5. A lawyer collects $750.75 ; what is his commission at 
3| %? Ans. $28.15+. 

6. My agent in New York has sold 3500 bushels of Indiana 
wheat @ $1.40, and 3600 bushels of dent corn @ $.74; what is 
his commission at 2} % ? 

7. A dealer in Philadelphia sells hides on commission of 8 J %, 
as follows: 2000 lb. Orinoco @ $.23J, 5650 lb. Central Ameri- 
can @ $.22, 450 lb. Texas @ $.23, and 650 lb. city slaughter 
@ $.21; what does he receive for his services ? Ans. $162.75. 

8. A commission merchant sold a consignment of flour and pork 
for $25372. He charged $132 for storage, and 6i % commis- 
sion ; what were the net proceeds of the sale ? 

9. An agent for a Rochester nurseryman sells 4000 apple trees 
at $25 per hundred, 2000 pear trees at $50 per hundred, 1600 
peach trees at $20 per hundred, 1800 cherry trees at $50 per 
hundred, and 500 plum trees at $50 per hundred; what is his 
commission at 30 %, and how much should he return to his em- 
ployer as the net proceeds, after deducting $203.50 for expenses ? 

Ans. Commission, $1041 ; Net proceeds, $2225.50. 

10. A lawyer having a debt of $785 to collect, compromises 
for 82 % ) what is his commission, at 5 % ? Ans. $32,185. 

11. 1 purchased in Chicago 4000 bushels of wheat @ $1.25, 
and shipped the same to my agent in Oswego, N. Y., who sold 
it @ $1.50 ; how much did I make, after paying expenses amount- 
ing to $415, and a commission of 3 % ? Ans. $405. 

12. An agent received $63 for collecting a debt of $1260 ; 
what was the rate of his commission ? Ans. 5 % . 

13. My Charleston agent has charged $74.25 for purchasing 
26400 lb. of rice at $4.50 per 100 lb. ; required the rate of his 
commission. 

14. A house and lot were sold for $7850, and the owner re- 



COMMISSION. 271 

ceived $7732.25 as the net proceeds; what was the rate of com- 
mission ? 

15. A commission merchant in Boston having received 28000 
lb. of Mobile cotton, effects a sale at $.12 \ per pound. After 
deducting $35.36 for freight and cartage, $10.50 for storage, and 
his commission, he remits to his employer $3252.89 as the net 
proceeds of the sale ; at what rate did he charge commission ? 

Arts. 5f %. 

16. The net proceeds of a sale were $5635, the commission was 
$115; what was the rate of commission ? 

17. An agent received $22.40 for selling grain at a commission 
of 4 <f ; what amount of grain did he sell ? Arts. $560. 

18. My attorney, in collecting a note for me at a commission of 
8 %, received as his fee $6.80 ; what was the face of the note ? 

19. Sent to my agent in Boston $245, to be invested in French 
prints at $.15 per yard, after deducting his commission of 2 % ; 
how many yards shall I receive ? Arts. 1666|. 

20. John Kennedy, commission merchant, sells for Ladd & Co. 
860 barrels of flour @ $7.50, on a commission of 2 $ %. He 
invests the proceeds in dry goods, after deducting his commission 
of 1 J % f° r purchasing ; how many dollars' worth of goods do 
Ladd & Co. receive? Ans. $6195.81. 

21. A commission merchant, whose rate both for selling and 
investing is 5 <f , receives 24000 lbs. of pork, worth 6 cents, and 
$3000 in cash, with instructions to invest in a shipment of cotton 
to London. What will be his entire commission ? Ans, $280. 

22. A speculator received $3290 as the net proceeds of a sale, 
after allowing a commission of 6 °J ) what was the value of the 
property? Ans. $3500. 

23. The net proceeds of a shipment of 500 tons of pressed hay, 
after deducting a commission of 3 % , and $500 for other charges, 
were $6290 ; what was the selling price per ton ? 

24. I send a quantity of dry goods into the country to be sold 
at auction, on commission of 9 % • What amount of goods must 
be sold, that my agent may buy produce with the avails, to the 
value of $3500, after retaining his purchase commission of 4 % ? 



272 PERCENTAGE 

Note 2. — $3500 plus the agent's commission equals the net proceeds of the 
sale. 

25. Having sold a consignment of cotton on 3 % commission, 
I am instructed to invest the proceeds in city lots, after deducting 
my purchase commission of 2 <f . My whole commission is $265; 
what is the cost of the city lots? Ans. $5141. 

26. What tax must be assessed to raise $50000, the collector's 
commission being f % ? Ans. $50377.83+. 

STOCKS. 

4:03. A Company is an association of individuals for the 
prosecution of some industrial undertaking. Companies may be 
incorporated or unincorporated. 

464. A Corporation is a body formed and authorized by law 
to act as a single person. 

465. A Charter is the legal act of incorporation, and defines 
the powers and obligations of the incorporated body. 

460. A Firm is the name under which an unincorporated 
company transacts business. 

Note. — A private banking company, or a manufacturing or commercial firm 
is also called a House. 

467. The Capital Stock of a corporation is the money con- 
tributed and employed to carry on the business of the company. 

468. Joint Stock is the money or capital of any company, 
incorporated or unincorporated. 

469. Scrip or Certificates of Stock are the papers or docu- 
ments issued by a corporation, giving the members their respective 
titles or claims to the joint capital. 

470. A Share is one of the equal parts into which capital 
stock is divided. The value of a share in the original contribu- 
tion of capital varies in different companies; in bank, insurance, 
and railroad companies of recent organization, it is usually $100. 

471. Stockholders are the owners of stock, either by original 
title or by subsequent purchase. The stockholders constitute the 
company. 

Notes. — 1. The capital stock of any corporation is limited by the charter. As 
a general rule, only a portion is paid at the time of subscription, the residuo 
being reserved for future outlays or disbursements. 



stocks. 273 

2. When the capital stock has been all paid in, money may be raised, if neces- 
sary, by loans, secured by mortgage upon the property. The bonds issued for 
these loans entitle the holders to a fixed rate of interest. 

3. Stocks, as a general name, applies to the scrip and bonds of a corporation, 
to government bonds and public securities, and to all paper representing joint 
capital or claims upon corporate bodies. 

4. The members of an incorporated company are individually liable for the 
debts and obligations of the company, to the amount of their interest or stock 
in the company, and to no greater amount. But the members of a firm or house 
are individually liable for all the debts and obligations of the company, without 
regard to the amount of their share or interest in the concern. 

The calculations of percentage in stocks are treated in this work 
under the heads of 

Stock-jobbing, Assessments and Dividends, and Stock Invest- 
ments. 



STOCK-JOBBING. 

472. Stock-jobbing is the buying and selling of stocks with 
a view to realize gain from their rise and fall in the market. 

473. The Nominal or Par value of stock is the sum for 
which the scrip or certificate is issued. 

474. The Market or Real value of stock is the sum for 
which it will sell. 

475. Stock is At Par when it sells for its first cost, or 
nominal value. 

47©. Stock is Above Par, at a premium or advance, when it 
sells for more than its nominal value. 

477. Stock is Below Par, or at a discount, when it sells for 
less than its nominal value. 

Note. — When the business of a company pays large profits to the stock- 
holders, the stock will be worth more than its original cost; but when the busi- 
ness does not pay expenses, the value of the stock will be less than its original 
cost. The average market value of stock generally varies directly as the rate 
of profit which the business pays. 

478. A Stock Broker is a person who buys and sells stocks, 
either for himself, or as the agent of another. 

Note. — A person employed by a manufacturer, wholesale dealer, or commission 
merchant, to seek customers and close bargains, at or from his place of business, 
is called a broker, of the class or kind corresponding to his business. 

479. Brokerage is the fee or compensation of a broker. 

480. The calculations in stock-jobbing are based upon the 
following relations : 

s 



274 PERCENTAGE. 

I. Premium, discount, and brokerage are each a percentage, 
computed upon the par value of the stock as the base. 

II. The market value of stock, or the proceeds of a sale, is the 
amount or difference, according as the sum is greater or less than 
the par value. 

Note 1. — In all examples relating to stocks, $100 will be considered a share, 
unless otherwise stated. 

EXAMPLES FOR PRACTICE. 

1. What cost 54 shares of Reading Railroad stock, at 4£ °J 

premium ? 

operation. , Analysis. We first 

$5400 X .045 = $243, premium. compute the premium 

$5400 + $243 = 5643, Ans. upon the par value of the 

Or, stock, and find it to be 

5400 X $1,045 = $5643, Ans. $243 ; adding this to the 

$5400, we obtain the cost, 
or market value, $5643. Or, since every dollar of the stock will cost 
$1 plus the premium, or $1,045, $5400 will cost 5400 X $1,045 = 
$5643. 

2. What do I receive for 32 shares of telegraph stock, which a 
broker sells for me at 15 % discount charging \ cf brokerage ? 

operation. Analysis. Adding 

.15 + .0025 = 1525 the rate of brokera S e to 

$L00 — $1525 = $.8475 proceeds the rate of discount > we 

of $1 of stock. have - 1525 ; hence $1 

3200 X $.8475 = $2712, Ans. will bring $1— $1525= 

$.8475, and $3200 will 
bring 3200 X $.8475 = $2712. 

3. I put $35400 into the hands of a broker to be invested in 
Missouri State Bonds, when their market value is 12 % below par; 
how many shares shall I receive, if the broker charges i % for 
his services ? 

OPERATION. 

$1.00 — $12 = $.88, market value of $1. 
$ .88 + $.00 J = .885, cost of $1. 
$35400 -r- .885 = $40000 = 400 shares, Ans. 
Analysis. Since the stock is 12 °/ below par, the market value of 
$1 is $.88 ; adding the rate of brokerage, we find that every dollar of 



stocks. 275 

the stock will cost me $.885. Hence for $3500 the broker can buy 

$35400 4- .885 = $40000 = 400 shares. 

Notes. — 2. The rate of brokerage in New York city has been fixed by cus- 
tom at i per cent. 

3. Since brokerage as the same base has the premium or discount, the rate of 
brokerage may always be combined with the rate of premium or discount, by 
addition or subtraction, as the nature of the question may require. 

4. The price of stock is usually quoted at a certain per cent, of the face, or 
nominal value. Thus stock at 4 <jL above par is quoted at 104 <b ; stock at 5 % 
below par is quoted at 95 fa ; and so on. 

4. What is the market value of 15 Ohio State bonds at 112 <f ? 

Ans. $1680. 

5. What shall I realize on 20 shares of Panama railroad stock 
at 135 (f , brokerage at If % ? Ans. $2665. 

6. My agent bought for me 120 shares of N. Y. Central rail- 
road stock, paying 80 f %, and charging brokerage at \ % ) what 
did the stock cost me ? Ans. $9750. 

7. What cost 36 shares in the Merchants' Bank, at a premium 
of 7 J %, brokerage \ %? 

8. A speculator invested $21915 in shares of the Harlem rail- 
road, at a discount of 601 % ) how many shares did he buy ? 

9. If 400 shares of the Bank of Commerce sell for $40150, 
what is the rate of premium? ^ Ans. f <f . 

10. A broker receives $48447 to be invested in bonds of the 
Michigan Central railroad, at 94i % ; how much stock can he 
buy, allowing \\ <f brokerage? 

11. My agent sells 830 barrels of Oenesee flour at $6 per barrel, 
commission 5 %, and invests the proceeds in stock of the Penn- 
sylvania Coal Company, at 82f %, charging \ °f for making the 
purchase; how many shares do I receive? Ans. 57. 

12. I purchased 18 shares of Ocean Telegraph stock, par value 
$500 per share, at a premium of 2 <f € , and sold the same at a dis- 
count of 28 % ; what was my loss ? Ans. $2700. 

' Note 5.— The rate of loss is .02. + .28 = .30, or .30 %. 

13. A speculator exchanged $3600 of railroad bonds, at 5 % 
discount, for 27 shares of stock of the Suffolk Bank, at 3 % 
premium, receiving the difference in cash ; how much money did 
he receive? 

14. A merchant owning 525 shares in the American Exchange 



276 



PERCENTAGE. 



Bank, worth 104 %, exchanges them for United States bonds 
worth 105 % ; how much of the latter stock does he receive? 

15. I purchased 12 shares of stock at a premium of 5 %, and 
sold the same at a loss of $96 ; what was the selling price ? 

16. Having bought $64000 stock in the Cunard Line, a 2 <f 
premium, at what price must I sell it, to gain $2560 ? 

Ans. 106 %. 

17. A speculator bought 250 shares in a Carson Valley mining 
company at 103 <f 0} and 150 shares of the Western Railroad stock 
at 95 °Jo ) h e exchanged the whole at the same rates, for shares in 
the N. Y. Central Railroad at 80 %, which he afterward sold at 
85 %. How much did he gain? Ans. $2500. 

18. I purchased stock at par, and sold the same at 3 % pre- 
mium, thereby gaining $750 ; how many shares did I purchase ? 

19. A broker bought Illinois State bonds at 103 %, and sold 
at 105 %. His profits were $240; what was the amount of his 
purchase? Ans. $12000. 

20. A man invested in mining stock when it was 4 % above 
par, and afterward sold his shares at 5 J °/ discount. His loss 
in trade was $760; how many shares did he purchase? 

21. I invested $6864 in Government bonds at 106 %, paying 
\\ <f brokerage, and afterward sold the stock at 112 %, paying 
1} % brokerage; what was my gain? Ans. $208. 

22. How much money must be invested in stocks at 3 % ad- 
vance, in order to gain $480 by selling at 7 % advance ? 

23. How many shares of stock must be sold at 4 % discount, 
brokerage i %, to realize $4775 ? Ans. 50. 

INSTALMENTS, ASSESSMENTS, AND DIVIDENDS. 

4L81. An Installment is a portion of the capital stock re- 
quired of the stockholders, as a payment on their subscription. 

4.82. An Assessment is a sum required of stockholders, to 
meet the losses or the business expenses of the company. 

483. A Dividend is a sum paid to the stockholders from the 
profits of the business. 



stocks. 277 

484. Gross Earnings are all the moneys received from the 
regular business of the company. 

485. Net Earnings are the moneys left after paying expenses, 
losses, and the interest upon the bonds, if there be any. 

48 G. In the division of the net earnings, or the apportion- 
ment of dividends and assessments, the calculations are made by 
finding the rate per cent, which the sum to be distributed or as- 
sessed bears to the entire capital stock. Hence, 

487. Dividends and assessments are a percentage computed 
upon the par value of the stock as the base. 

EXAMPLES FOR PRACTICE. 

1. The Long Island Insurance Company declares a dividend 
of 6 % ; what does A receive, who owns 14 shares ? 

Analysis. , According to 

operation. 449, we multiply the base, 

$1400 X .06 = $84 $1400, by the rate, .06, and 

obtain the dividend, $84. 

2. A canal company whose subscribed funds amount to $84000, 
requires an instalment of $6300 j what per cent, must the stock- 
holders pay? 

operation. Analysis. According to 

$6300 -f- 84000 = .07i 450, we divide the in- 

stalment, $6300, which is 
percentage, by the base, $84000, and obtain the rate, .07 J = 7 J % . 

3. A man owns 56 shares of railroad stock, and the company 
has declared a dividend of 8 % ; what does he receive ? 

Arts. $448. 

4. I own $15000 in a mutual insurance company ; how many 
shares shall I possess after a dividend of 6 % has been declared, 
payable in stock ? Ans. 159 shares. 

5. The Pittsburgh Gas Company declares a dividend of 15 % ; 
what will be received on 65 shares ? 

6. A received $600 from a 4 % dividend ; how much stock 
did he own? Ans. $15000. 

24 



278 PERCENTAGE. 

7. The paid-in capital «of an insurance company is $536000. 
Its receipts for one year are $99280, and its losses and expenses 
are $56400; what rate of dividend can it declare? Ans. 8 %. 

8. The net earnings of a western turnpike are $3616, and the 
amount of stock is $56000 ; if the company declare a dividend 
of 6 </o, what surplus revenue will it have? Ans. $256. 

9. The capital stock of the Boston and Lowell Railroad Co. 
is $1830000, and its debt is $450000. Its gross earnings for the 
year 1858 were $407399, and its expenses $217621. If the com- 
pany paid expenses, and interest on its debt at 5| %, what divi- 
dend would a stockholder receive who owned 30 shares ? 

Ans. $240. 

10. The charter of a new railroad company limits the stock to 
$800,000, of which 3 instalments of 10 %, 25 %, and 35 %, re- 
spectively, have been already paid in. The expenditures in the con- 
struction of the road have reached the sum of $540,000, and the 
estimated cost of completion is $400,000. If the company call in 
the final instalment of its stock, and assess the stockholders for the 
remaining outlay, what will be the rate % ? Ans. 17 J. 

11. The Bank of New York, having $156753.19 to distribute 
to the stockholders, declares a dividend of 5i % ; what is the 
amount of its capital ? Ans. $2,985,775. 

12. The passenger earnings of a western railroad in one year 
were $574375.25, the freight and mail earnings were $643672.36, 
the whole amount of disbursements were $651113.53, and the 
company was able to declare a dividend of 8 % ; how much scrip 
had the company issued? Ans. $7086676. 

13. Having received a stock dividend of 5 %, I find that I 
own 504 shares ; how many shares had I at first ? Ans. 480. 

14. I received a 6 % dividend on Philadelphia City railroad 
stock, and invested the money in the same stock at 75 % . My 
stock had then increased to $16200 ; what was the amount of my 
dividend? Ans. $900. 

15. A ferry company, whose stock is $28000, pays 5 % divi- 
dends semi-annually. The annual expenses of the ferry are 
$2950 ; what are the gross earnings ? Ans. $5750. 



stocks. 279 

stock investments. 

48 8. The net earnings of a corporation are usually divided 
among the stockholders, in semi-annual dividends. The income 
of capital stock is therefore fluctuating, being dependent upon 
the condition of business ; while the income arising from bonds, 
whether of government or corporations, is fixed, being a certain 
rate per cent., annually, of the par value, or face of the bonds. 

4:80. Stock producing a regular income receives a designation 
corresponding to the rate. Thus, bonds drawing 6 % annually 
are called 6 per cent, stock, or 6's ; stocks yielding 7 % are called 
7's; and so on. 

Note. — The profitableness of an investment depends jointly upon the price 
paid for the stock, and the rate of income which the stock produces. 

CASE I. 

490. To find what income any investment will pro- 
duce. 

1. What income will be obtained by investing $6840 in stock 
bearing 6 %, and purchased at 95 % ? 

operation. Analysis. We di- 

$6840 -4- .95 = $7200, stock purchased. vide the investment, 

$7200 x .06 == $432, annual income. $6840, by the cost of 

$1, and obtain $7200, 
the stock which the investment will purchase, (452). And since the 
stock bears 6 fo interest, we have $7200 X .06 == $432, the annual 
income obtained by the investment. Hence, 

Rule. Find how much stock the investment will purchase, and 
then compute the income at the given rate upon the par value. 

EXAMPLES FOR PRACTICE. 

1. The trustees of a school invested $35374.80 in the IT. S. 
5 °Jo bonds as a teachers' fund, purchasing the stock at 102 J % ; 
if the salary of the Principal be $1000, what sum will be left to 
pay assistants ? Ans. $725.60. 

2. A young man receiving a legacy of $48000, invested one 
half in 5 % stock at 95 \ %, and the other half in 6 % stock at 



280 PERCENTAGE. 

112 %, paying brokerage at i % ; what annual income did he 
secure from his legacy ? Ans. $2530. 

3. A capitalist holding bonds of the Illinois Central Railroad 
to the amount of $90000, exchanged them at the market price of 
88 (foj for capital stock in the same company, worth 62 i <f . The 
bonds drew 7 °Jo annually, while the stockholders received two 
dividends during the year, the first of 3 %, and the second of 
3 J % ; how much did the capitalist gain annually by the ex- 
change ? Ans. 1936.80. 

5. I have $32300 to invest, and can buy New York Central 6's 
at 85 %, or New York Central 7's at 95 % ; how much more 
profitable will the latter be than the former, per year ? 



CASE II. 

491. To find what sum must be invested, to obtain a 
given income. 

I. What sum must be invested in Virginia 5 per cent, bonds, 
purchasable at 80 %, to obtain an income of $600 ? 

operation. Analysis. Since 

$600 -+- .05 = $12000, stock required. H of the stock will 

$1200 X .80 = $9600, cost, or investment, obtain $.05 income, 

to obtain $600 will 
require $600 4- .05 = $12000, (Case I). Multiplying the par value 
of the stock by the market price of $1, we have $12000 X .80 == 
$9600, the cost of the required stock, or the sum to be invested. 
Hence the 

Rule. I. Divide the given income by the % which the stock 
pays ; the quotient will be the par value of the stock required. 

II. Multiply the par value of the stock by the market value of 
one dollar of the stock ; the product will be the required investment. 

EXAMPLES FOR PRACTICE. 

1. What sum must I invest in the Michigan Central 8 per 
cents., selling at 85J %, to secure an annual income of $1200? 

Ans. $12825. 



STOCKS. 281 

2. If Missouri State 6's are 16 % below par, what sum must 
be invested in this stock to obtain an income of $960 ? 

3. I have an investment in Virginia 6 per cent, bonds, which 
brings me an income of $630, and I wish to sell, and invest the 
proceeds in United States 5 per cents. ; if in the market, the Vir- 
ginia bonds are 15 % below par, and the United States bonds 3 °J 
above par, what sum must I add to the investment to secure the 
same income ? Arts. $4053. 

4. How many shares of canal stock at 60 % must be sold, in 
order that the proceeds, invested in the California State 7's at 
96 <f 0y may yield an income of 840 ? Ans. 192 shares. 

5. A capitalist invested equal sums in the United States 5 per 
cents., and the Maryland 5 per cents., purchasing the former at a 
premium of 4 %, and the latter at a discount of 5 cf . The in- 
come from both investments was $2487.50; how much money was 
invested in each kind of stock ? Ans. $24700. 

CASE III 

492. To find what per cent, the income is of the in- 
vestment, when stock is purchased at a given price. 

1. What per cent, of my investment shall I secure, by pur- 
chasing the New York 7 per cents, at 105 <f ? 

Analysis. Since $1 of the 

operation. stock will cost $1.05, and pay 

.07 ~ 1.05 = 6f % $.07, the income is T J T = 6| <f 

of the investment. Hence the 

Rule. Divide the annual rate of income which the stock hears 
by the price of the stock ; the quotient will be the rate upon the in- 
vestment. 

EXAMPLES FOR PRACTICE. 

1. What per cent, of his money will a man obtain, by investing 
in 6 per cent, stock at 108 % ? Ans. 5| %. 

2. What is the rate of income upon money invested in 6 per 
cent, bonds, purchased at a discount of 16 °J ? Ans. 7^ %. 

3. Panama railroad stock is at a premium of 34? <f 0) and the 

24* 



282 PERCENTAGE. 

charge for brokerage is 1} % ; what will be the rate of income 
on an investment in these funds, if the stock pays a dividend of 
8 J % annually? Arts. 6 I </ . 

4. Which is the better investment, to buy 5's at 70 %, or 6's 
at 80 % ? 

5. Which is the more profitable, to buy 8's at 12 ^, or 5's at 
75 % ? 

6. If a railroad company, whose capital stock is $24182400, 
pays $1722996 in a dividend to the stockholders during a certain 
year, what per cent, on the investment will a man receive who 
holds shares purchased at 79| % ? Arts. 8{f %. 

CASE IV. 

493. To find the price at which, stock must be pur- 
chased, to obtain a given rate upon the investment. 

1. At what price must 6 per cent, stock be purchased, in order 
to obtain 8 per cent, income on the investment ? 

operation. Analysis. Since $.06, the in- 

$.06 -*- .08 = $75 come of $1 of the stock, is 8 <f of 

the sum paid for it, we have, 
( 449 ), $.06 -=- .08 = $.75, the purchase price. Hence, 

Rule. Divide the annual rate of income which the stock bears 
by the rate required on the investment) the quotient will be the 
price of the stock. 

EXAMPLES FOR PRACTICE. 

1. What must I pay for Government 5 per cents., that my in- 
vestment may yield 8 % ? Ans. 62 $ c/ . 

2. At what rate of discount must the Vermont 6 per cent, bonds 
be purchased, that the person investing may secure 6£ % upon 
his money? Ans. 4 %. 

3. What rate of premium does 7 per cent, stock bear in the 
market, when an investment pays Q % ? 

4. A speculator invested in a Life Insurance Company, and re- 
ceived a dividend of 6 %, which was 8 J % on his investment; 
at what price did he purchase? Ans. 72 %. 



PROFIT AND LOSS. 9g3 

PROFIT AND LOSS. 

494. Profit and Loss are commercial terms/used to express 
the gain or loss in business transactions. , 

4«J*S. Gains and losses are usually estimated by some rate per 
cent, on the money first expended or invested. Hence 

I. Profit and loss are reckoned as percentage upon the prime or 
first cost of the goods as the base. 

II. The selling price of the goods is amount or difference, ac- 
cording as it is greater or less than the prime cost. 

EXAMPLES FOR PRACTICE. 

1. A merchant bought cloth for $3.25 per yard, and gained 
8 % in selling ; what was the selling price ? 

operation. Analysis. Multi- 

$3.25 X .08 = $.26, advance in price. P^g the prime 

$3.25 + .26 = $3.51, selling price. cost, $3.25, which is 

Q r the base of gain, 

$3.25 x 1.08 = $3.51, selling price. £>' h ? , the ™% 

. .08, we have $.26, 

the gain, which added to the cost gives $3.51, the selling price. Or, 

since the rate of gain is 8 %, that which cost $1 will bring $1.08, 

and the selling price will be 1.08 times the buying price. Hence 

$3.25 X 1.08 = 3.51, the selling price. 

2. A jobber invested $2560 in dry goods, and realized $384 
net profit ; what was the rate per cent, of his gain ? 

operation. Analysis. According to 

$384 -h $2560 = 15 % (Prob. II, 460), we divide the 

gain, $384, which is percent- 
age, by the cost, $2560, which is the base, and obtain 15 = 15 % , the 
rate of gain. 

. 3. A produce dealer sold a shipment of wheat at a loss of 5 %, 
realizing as the net proceeds, $8170 ; what was the cost ? 

operation. Analysis. According to 

$1.00 — .05 = .95 (Prob. Y, 453), we divide the 

8170 -*- .95 = $8600, Ans. net proceeds, $8170, which 

is difference, (448), by 1 
minus the rate of loss, or .95, and obtain the base, or prime cost, \ 



284 PERCENTAGE. 

4. A merchant pays $7650 for a stock of spring goods ; if lie 
sell at an advance of 20 % upon the purchase price, what will be 
his profits, after deducting $480 for expenses? Am. $1050. 

5. Bought 320 yards of calico @ 15 cents, and sold it at a 
reduction of 2 J % ; what was the entire loss? 

6. A dealer having bought 30 barrels of apples at $3.50 per 
barrel, and shipped them at an expense of $5.38, to be sold on 
commission of 5 %, what will be his whole loss if the selling 
price is 10 % below the purchase price ? Am. $20.60^. 

7. Bought corn at $.50 a bushel; at what price must it be sold 
to gain 33J per cent. ? 

8. Bought fish at $4.25 per quintal, and sold the same at $4.93 ; 
what was my gain per cent. ? Am. 16 %. 

9. Bought a hogshead of sugar containing 9 cwt. 44 lb., for $59 ; 
paid $4.72 for freight and cartage ; at what price per pound must 
it be sold to gain 20 per cent, on the buying price ? 

10. A wine merchant bought a hogshead of wine for $157.50 ; 
a part having leaked out he sold the remainder for $3.32 J a gallon, 
and found his loss to be 5 per cent, on the cost ; how many gallons 
leaked out? Am. 18. 

11. Sold a farm of 106 A. 3 R. 30 P. for $96 an acre, and 
gained 18 per cent, on the cost; how much did the whole farm 
cost? Am. $8700. 

12. A lumberman sold 36840 feet of lumber at $21.12 per M, 
and gained 28 per cent. ; how much would he have gained or lost, 
had he sold it at $17 per M? Am. $18.42, gained. 

13. A speculator bought shares in a mining company when the 
stock was 4 % below par, and sold the same when it was 28 <f 
below par ; what per cent, did he lose on his investment ? 

14. A machinist sold a fire engine for $7050, and lost 6 per 
cent, on its cost; for how much ought he to have sold it to gain 
12 J per cent.? Am. $8437.50. 

15. Sold my carriage at 30 per cent, gain, and with the money 
bought another, which I sold for $182, and lost 12^ per cent.; 
how much did each carriage cost me ? * f First, $160 ; 

US ' 1 Second, $208. 



PROFIT AND LOSS. 285 

16. Gafney, Burke & Co. bought a quantity of dry goods for 
$6840; they sold J of them at 15 per cent, profit, i at 18f per 
cent., i at 20 per cent., and the remainder at 33 J per cent, profit; 
how much was the average gain per cent., and how much the whole 
gain ? Ans. 21| % gain ; $1482, entire gain. 

17. If I buy a piece of land, and it increases in value each 
year at the rate of 50 per cent, on the value of the previous year, 
for 4 years, and then is worth $12000, how much did it cost ? 

18. A Western merchant bought wheat as follows : 600 bushels 
of red Southern @ $1.80, 1200 bushels of white Michigan @ 
$1.62i, and 200 bushels of Chicago spring, @ $1.25. He 
shipped the whole to his correspondent in Buffalo, who sold the 
first two kinds at an advance of 20 % in the price, and the bal- 
ance at $1.20 per bushel, and deducting from the gross avails his 
commission at 5 %, and $220.40 for expenses, returned to the 
consignor the net proceeds. What was the rate of the merchant's 
gain? Ans. 4 J %. 

19. A broker buys stock when it is 20 % below par, and sells 
it when it is 16 % below par ; what is his rate of gain ? 

20. A man has 5 per cent, stock the market value of which is 
78 % ; if he sells it, and takes in exchange 6 % stock at 4 % 
premium, what per cent, of his annual income does he lose ? 

21. A machinist sold 24 grain-drills for $125 each. On one 
half of them he gained 25 per cent., and on the remainder he 
lost 25 per cent.; did he gain or lose on the whole, and how 
much ? Ans. Lost 200. 

22. Bought land at $30 an acre; how much must I ask an acre, 
that I may abate 25 per cent, from my asking price, and still make 
20 per cent, on the purchase money ? Ans. $48. 

23. A salesman asked an advance of 20 per cent, on the cost 
of some goods, but was obliged to sell at 20 per cent, less than 
his asking price ; did he gain or lose, and how much per cent. ? 

24. A Southern merchant ships to his agent in Boston, a quan- 
tity of sugar consisting of 200 bbl. of New Orleans, each containing 
216 lb., purchased at 5 cents per pound, and 560 bbl. of West 
India, each containing 200 lb., purchased at 5f cents per pound. 



286 PERCENTAGE. 

The agent's account of sales shows a loss of 1 % on the New Or- 
leans, and a profit of || °J on the "West India sugar; does the 
merchant gain or lose on the whole consignment, and what per 
cent.? Arts. Gains § °/ - 

25. A grocer sold a hogshead of molasses for $31.50, which 
was a reduction of 30 % from the prime cost ; what was the pur- 
chase price paid per gallon ? 

26. A speculator sold stock at a discount of 7| %, and made 
a profit of 5 % ; at what rate of discount had he purchased the 
stock? Ans. 12 %. 

27. A dry-goods merchant sells delaines for 2| cents per yard 
more than they cost, and realizes a profit of 8 % ; what was the 
cost per yard ? Ans. $.31i. 

28. If I make a profit of 18| °J by selling broadcloth for $.75 
per yard above cost, how much must I advance on this price to 
realize a profit of 31| % ? 

29. A speculator gained 30 % on | of his investment, and lost 
5 c/o on the remainder, and his net profits were $720. What 
would have been his profits, had he gained 30 % on | and lost 
5 % on the remainder? Ans. $405. 

30. A man wishing to sell his real estate asked 36 per cent, 
more than it cost him, but he finally sold it for 16 per cent, less 
than his asking price. He gained by the transaction $740.48. 
How much did the estate cost him, what was his asking price, and 
for how much did he sell it ? 

Ans. Cost, $5200 ; asking price, $7072; sold for $5940.48. 

31. Sold | of a barrel of beef for what the whole barrel cost; 
what per cent, did I gain on the part sold ? 

32. Bought 4 hogsheads of molasses, each containing 84 gal- 
lons, at $.37} a gallon, and paid $7.50 for freight and cartage. 
Allowing 5 per cent, for leakage and waste, 4 per cent, of the sales 
for bad debts, and 1 per cent, of the remainder for collecting, for 
how much per gallon must I sell it to make a net gain of 25 per 
cent, on the cost of the whole? Ans. $.55 — . 



INSURANCE. . 287 

INSURANCE. 

496. Insurance is security guaranteed by one party to ano- 
ther, against loss, damage, or risk. It is of two kinds ) insurance 
on property, and insurance on life. 

497. The Insurer or Underwriter is the party taking the 
risk. 

498. The Insured or Assured is the party protected. 

499. The Policy is the written contract between the parties. 
tSOO. Premium is the sum paid for insurance. It is always a 

certain per cent, of the sum insured, varying according to the 

degree or nature of risk assumed, and payable annually or at stated 

intervals. 

Notes. — 1. Insurance business is generally conducted by joint stock compa- 
nies, though sometimes by individuals. 

2. A Mutual Insurance company is one in "which each person insured is enti- 
tled to a share in the profits of the concern. 

3. The act of insuring is sometimes called taking a risk. 

FIRE AND MARINE INSURANCE. 

501 . Insurance on property is of two kinds ; Fire Insurance 
and Marine Insurance. 

Fire Insurance is security against loss of property by fire. 

Marine Insurance is security against the loss of vessel or cargo 
by the casualties of navigation. 

o©5§. The Sum Covered by insurance is the difference be- 
tween the sum insured and the premium paid. 

Notes. — 1. As security against fraud, most insurance companies take risks at 
not more than two- thirds of the full value of the property insured. 

2. When insured property suffers damage less than the amount of the policy, 
the insurers are required to pay only the estimated loss. 

503. The calculations in insurance are based upon the fol- 
lowing relations : 

I. Premium is percentage (4t4U5) • 

II. The sum insured is the base of premium. 

III. The sum covered by insurance is difference. 

EXAMPLES FOR PRACTICE. 



1. What premium must be paid for insuring my stock of goods 
to the amount of $5760 at 1} % ? 



288 PERCENTAGE. 

operation. Analysis. According to 

$5760 X .0125 = $72, Ans. Prob. I, (449), we multiply 

$5760, the base of premium, 
by .0125, the rate, and obtain $72, the premium. 

2. For what sum must a granary be insured at 2 % in order to 
cover the loss of the wheat, valued at $1617 ? 

operation. Analysis. According to 

1.00 .02 = .98 Pr0D ' V> (453), we divide the 

$1617 -f- .98 = $1650, Ans. sum to be covered, $1617, 

which is difference, by 1 
minus the rate of premium, and obtain $1650, the base of premium, 
or the sum to be insured. 

Proof. $1650 X .02 = $33, premium ; $1650 — $33 = $1617, the 
sum covered. 

3. What must be paid for an insurance of $5860 at 1J % ? 

4. What is the premium of $860 at $ % ? Ans. $4.30. 

5. What is the premium for an insurance of $3500 on my house 
and barn, at 1 1 % ? Ans. $43.75. 

6. A fishing craft, insured for $10000 at 2 J %, was totally 
wrecked; how much of the loss was covered? Ans. $29775. 

7. A hotel valued at $10000 has been insured for $6000 at 
li %, $5.50 being charged for the policy and the survey of the 
premises ; if it should be destroyed by fire, what loss would the 
owner suffer? Ans. $4080.50. 

8. A merchant whose stock in trade is worth $12000, gets the 
goods insured for | of their value, at f % ; if in a conflagration 
he saves only $2000 of the stock, what actual loss will he sustain ? 

9. If I take a risk of $36000 at 2} %, and re-insure £ of it 
at 3 %, what is my balance of the premium ? Ans. $360. 

10. I pay $12 for an insurance of $800 ; what is the rate of 
premium? Ans. \\ cj . 

11. A trader got a shipment of 500 barrels of flour insured for 
80 % of its cost, at 3 J %, paying $107.25 premium; at what 
price per barrel did he purchase the flour? Ans. $8.25. 

12. The Astor Insurance Company took a risk of $16000, for 
a premium of $280 ; what was the rate of insurance ? 

13. A whaling merchant gets his vessel insured for $20000 in 



INSURANCE. 289 

the Gallatin Company, at f %, and for $30000 in the Howard 
Company, at J % ; what rate of premium does he pay on the whole 
insurance? Ans. § %. 

14. If it cost $46.75 to insure a store for J of its value, at If 
<f , what is the store worth ? Ans. $6800. 

15. For what sum must I get my library insured at 14 %, to 
cover a loss of $79110 ? Ans. $8000. 

16. What will be the premium for insuring at 2f %, to cover 
$27320? Ans. $680. 

17. A shipment of pork was insured at 4f %, to cover f of its 
value. The premium paid was $122.50; what was the pork 
worth? Ans. $4284. 

18. A gentleman obtained an insurance on his house for f of 
its value, at 1J % annually. After paying 5 instalments of pre- 
mium, the house was destroyed by fire, in consequence of which 
he suffered a loss of $2940 ; what was the value of the house ? 

19. A man's property is insured at 2 J % payable annually; 
in how many years will the premium equal the amount of the 
policy ? Ans. 40 years. 

20. A company took a risk at 2 J %, and re-insured | of it in 
another company at 2 J %. The premium received exceeded the 
premium paid by $72. What was the amount of the risk ? 

21. The Commercial Insurance Company issued a policy of 
insurance on an East India merchantman for f of the estimated 
value of ship and cargo, at 4J % , and immediately re-insured | 
of the risk in the Manhattan Company, at 3 %. During the out- 
ward voyage the ship was wrecked, and the Manhattan Company 
lost $1350 more than the Commercial Company; what did the 
owners lose? Ans. $40590. 

LIFE INSURANCE. 

504. Life Insurance is a contract in which a company agrees 
to pay a certain sum of money on the death of an individual, in 
consideration of' an immediate payment, or of an annual premium 
paid for a term of years, or during the life of the insured. The 

25 t 



290 PERCENTAGE. 

policy may be made payable to the heirs of the insured, or assured, 
person, or to any one whom he may designate. 

505. The policies issued by life insurance companies are of 
various kinds, the principal of which are as follows : 

1st. Term policies, payable on the death of the insured, if the 
death occurs during a specified term of years ; these require the 
payment of an annual premium till the policy matures or expires. 

2d. Life policies, payable on the death of the insured, the annual 
premium to continue during life. 

3d. Life policies, payable on the death of the insured, the annual 
premium to cease at a given age. 

4th. Endowment assurance policies, payable to the assured 

person on his attaining a given age, or to his heirs if his death 

occurs before that age, annual premium being required till the 

policy matures. 

Note. — The premium on the first and second classes of policies may be dis- 
charged by a single payment, instead of annual instalments. 

506. The Expectation of Life is the average number of years 
of life that remains to a person of a given age, as determined by 
tables of mortality. 

507. The rates of life insurance, as fixed by different com- 
panies, are based upon the expectation of life and the probable 
rates of interest which money will bear in future time. 

508. The rates of annual premium for the assurance of $100 
on a single life, according to the two kinds of life policies (2 and 
3), as issued by the Mutual Life Insurance Company of New 
York, are given in the Life Table on page 291. 

509. The rates of annual premium of an assurance of $100 
in the same company, payable to the party assured on his attain- 
ing the age of 40, 45, 50, 55, 60, or 65, or to his representatives, 
in case of death before attaining these ages respectively, are shown 
in the Endowment Assurance Table on page 292. 

Note.— The tables of the Mutual Life Insurance Company of New York have 
been selected, as furnishing good examples of a variety of policies ; the compu- 
tations by any other tables would not differ in any material respect from those 
introduced under these tables. 



INSURANCE. 
LIFE TABLE. 



291 



ANNUAL PREMIUM ON A POLICY OF $100. 



Age at 
issue. 


Payments 
during life. 


Payments 
To cease at 65. 


Payments 
To cease at 60. 


Payments 
To cease at 50. 


Age at 
issue. 


14 
15 


$1.4707 
1.5105 


$1.4999 
1.5422 


$1.5238 
1.5683 


$1.6150 

1.6681 


14 
15 


16 


1.5516 


1.5861 


1.6145 


1.7240 


16 


17 


1.5940 


1.6316 


1.6625 


1.7826 


17 


18 


1.6377 


1.6786 


1.7124 


1.8444 


18 


19 


1.6829 


1.7275 


1.7644 


1.9096 


19 


20 


1.7296 


1.7782 


1.8186 


1.9785 


20 


21 


1.7780 


1.8310 


1.8753 


2.0516 


21 


22 


1.8280 


1.8859 


1.9344 


2.1292 


22 


23 


1.8798 


1.9431 


1.9963 


2.2118 


23 


24 


1.9335 


2.0027 


2.0612 


2.3000 


24 


25 


1.9891 


2.0648 


2.1291 


2.3944 


25 


26 


2.0470 


2.1300 


2.2007 


2.4959 


26 


27 


2.1071 


2.1981 


2.2761 


2.6054 


27 


28 


2.1696 


2.2695 


2.3555 


2.7238 


28 


29 


2.2346 


2.3444 


2.4395 


2.8525 


29 


30 


2.3023 


2.4230 


2.5284 


2.9928 


30 


31 


2.3728 


2.5058 


2.6226 


3.1466 


31 


32 


2.4464 


2.5930 


2.7228 


3.3163 


32 


33 


2.5232 


2.6851 


2.8296 


3.5044 


33 


34 


2.6034 


2.7824 


2.9436 


3.7142 


34 


35 


2.6873 


2.8856 


3.0657 


3.9503 


35 


36 


2.7752 


2.9951 


3.1971 


4.2182 


36 


37 


2.8674 


3.1117 


3.3387 


4.5251 


37 


38 


2.9641 


3.2361 


3.4919 


4.8807 


38 


39 


3.0658 


3.3692 


3.6584 


5.2981 


39 


40 


3.1729 


3.5120 


3.8402 


5.7959 


40 


41 


3.2856 


3.6654 


4.0393 




41 


42 


3.4046 


3.8311 


4.2588 




42 


43 


3.5303 


4.0106 


4.5021 




43 


44 


3.6632 


4.2055 


4.7735 




44 


45 


3.8038 


4.4181 


5.0782 




45 


46 


3.9530 


4.6512 


5.4235 




46 


47 


4.1111 


4.9075 


5.8180 




47 


48 


4.2782 


5.1902 


6.2726 




48 


49 


4.4549 


5.5038 


6.8032 




49 


50 


4.6417 


5.8536 


7.4317 




50 


51 


4.8393 


6.2470 






51 


52 


5.0486 


6.6935 






52 


53 


5.2708 


7.2061 






53 


54 


5.5067 


7.8017 






54 


55 


5.7577 


8.5048 






55 



292 



PERCENTAGE. 
ENDOWMENT ASSURANCE TABLE. 



ANNUAL PREMIUM ON A POLICY OF $100. 


Age at 


Policy due 


Policy due 


Policy due 


Policy due 


Policy due 


Policy due 


Age at 


issue. 


at 40. 


at 45. 


at 50. 


at 55. 


at 60. 


at 65. 


issue. 


14 




$2,475 


$2,113 


$1,868 


$1,704 




14 


15 




2.587 


2.197 


1.935 


1.759 




15 


16 


$3,356 


2.707 


2.285 


2.004 


1.816 


$1,694 


16 


17 


3.545 


2.835 


2.379 


2.077 


1.876 


1.746 


17 


18 


3.752 


2.937 


2.478 


2.153 


1.939 


1.799 


18 


19 


3.978 


3.122 


2.585 


2.234 


2.004 


1.855 


19 


20 


4.228 


3.283 


2.698 


2.320 


2.073 


1.914 


20 


' 21 


4.504 


3.458 


2.819 


2.410 


2.145 


1.974 


21 


22 


4.812 


3.648 


2.949 


2.506 


2.220 


2.038 


22 


23 


5.156 


3.855 


3.089 


2.608 


2.300 


2.104 


23 


24 


5.544 


4.083 


3.239 


2.717 


2.384 


2.174 


24 


25 


5.985 


4.333 


3.402 


2.832 


2.473 


2.247 


25 


26 


6.489 


4.611 


3.578 


2.956 


2.567 


2.323 


26 


27 


7.082 


4.920 


3.770 


3.088 


2.666 


2.404 


27 


28 


. 7.752 


5.265 


3.979 


3.231 


2.772 


2.489 


28 


29 


8.558 


5.654 


4.208 


3.384 


2.884 


2.578 


29 


30 


9.526 


6.096 


4.461 


3.549 


3.004 


2.672 


30 


31 




6.601 


4.740 


3.728 


3.132 


2.772 


31 


32 




7.185 


5.051 


3.923 


3.268 


2.877 


32 


33 




7.866 


5.398 


4.135 


3.415 


2.989 


33 


34 




8.673 


5.789 


4.368 


3.573 


3.108 


34 


35 




9.641 


6.232 


4.624 


3.743 


3.235 


35 


36 






6.739 


4.906 


3.928 


3.370 


36 


37 






7.325 


5.221 


4.128 


3.515 


37 


38 






8.008 


5.572 


4.347 


3.670 


38 


39 






8.816 


5.967 


4.586 


3.837 


39 


40 






9.787 


6.415 


4.849 


4.017 


40 


41 








6.927 


5.139 


4.212 


41 


42 








7.518 


5.462 


4.424 


42 


43 








8.207 


5.822 


4.655 


43 


44 








9.022 


6.227 


4.908 


44 


45 








10.000 


6.686 


5.185 


45 


46 










7.210 


5.491 


46 


47 










7.813 


5.830 


47 


48 










8.515 


6.208 


48 


49 










9.343 


6.630 


49 


50 










10.332 


7.105 


50 


51 










11.536 


7.645 


51 


52 












8.265 


52 


53 












8.983 


53 


54 












9.826 


54 


55 












10.831 


55 



INSURANCE. 293 

EXAMPLES FOB PRACTICE. 

1. "What sum must a man pay annually to the Mutual Insurance 
of New York, for a life policy of $2500, his age being 23 years 
at the issue of the policy ? 

operation. Analysis. We multi- 

$2500 X .025232 = $63.08, Arts. ply the face of the policy, 

$2500, by the rate per 

cent, found opposite 23 years in the Life Table, expressed decimally, 

and obtain $63.08, the annual premium required. 

Note. — The examples which follow all refer to the rates given in the pre- 
ceding tables. 

2. A man at 30 years of age takes a policy for $2000, the pay- 
ments of premium to cease at 50 ; if he survives that age, how 
much more money will he receive from the company than he pays 
to the company? Ans. $743.02. 

3. What annual premium must a man pay during life, com- 
mencing at the age of 50, to secure $3000 at his death ? 

4. A gentleman at the age of 36 gets his life insured for $1500, 
premium to cease at the age of 60 ; if he dies at 52, how much 
more will his family receive than has been paid out in premiums ? 

5. A clergyman wishing to secure an income to his family after 
his death, had his life insured at the age of 54, in the sum of 
$3500, premium payable during life; his decease took place at 
the age of 72. How much more would have been saved to his 
family if he had taken, instead, a policy for the same amount, with 
payments of premium to cease at 65 ? Ans. $385.24. 

6. How much more premium will be required to secure an en- 
dowment of $1200 at 40, by taking out a policy at the age of 30, 
than if the policy be taken at 24 ? Ans. $26.45. 

7. A man 37 years old took an endowment assurance policy 
for $750, due at the age of 50, and died when 49 years old; how 
much more would his heirs have realized if he had taken a life 
policy for the same amount, with payment to cease at 50 ? 

8. A has his life insured at the age of 20, and B has his in- 
sured at the age of 30, each taking a life policy requiring annual 
payments of premium during life ; what will be the age of each 

25* 



294 PERCENTAGE. 

when the amount of premium paid shall exceed the face of his 
policy ? Ans. A, 77 years ; B, 73 years. 

9. What is the whole amount of premiums that must be paid 
to secure an endowment of $1000 at the age of 60, the policy 
being issued at the age of 45 ? Ans. 1069.76. 

10. A person at the age of 34 had his life insured in the sum 
of $600, the premium to cease at 50. When he died, there was 
a net gain to his family of $421.72 ; how many payments of pre- 
mium had he made ? Ans. 8. 

11. A gentleman obtained an insurance on his life at the age 
of 29, and died at the age of 40 ; the policy taken required an- 
nual payments of premium during life, and secured to his heirs 
$1849.12 more than the whole premium paid. Required the face 
of the policy. Ans. $2500. 

TAXES. 

510. A Tax is a sum of money assessed on the person or pro- 
perty of an individual, for public purposes. 

511. A Poll Tax is a certain sum required of each male citi-. 
zen liable to taxation, without regard to his property. Each person 
so taxed is called & poll. 

512. A Property Tax is a sum required of each person own- 
ing property, and is always a certain per cent, of the estimated 
value of his property. 

513. An Assessment Roll is a list or schedule containing the 
names of all the persons liable to taxation in the district or com- 
pany to be assessed, and the valuation of each person's taxable 
property. 

514:. Assessors are the persons appointed to prepare the as- 
sessment roll, and apportion the taxes. 

1. In a certain town a tax of $4000 is to be assessed. There 
are 400 polls to be assessed $.50 each, and the valuation of the 
taxable property, as shown by the assessment roll, is $950000 ; 
what will be the property tax on $1, and how much will be A's 
tax, whose property is valued at $3500, and who pays for 3 polls ? 



TAXES. 



295 



OPERATION. 

$ .50 x 400 = $200, amount assessed on the polls. 
$4000 — $200 = $3800, amount to be assessed on property. 
$3800 ~- $95000 = .004, rate of taxation ; 
$3500 x .004 = $14, A's property tax; 
$ .50 x 3 == 1.50 , A's poll tax; 

$15.50, amount of A's tax. Hence the 
Rule. I. Find the amount of poll tax, if any, and subtract it 
from the whole tax to be assessed ; the remainder will be the prop- 
erty tax. 

II. Divide the property tax by the whole amount of taxable 
property ; the quotient will be the rate of taxation. 

III. Multiply each man's taxable property by the rate of taxa- 
tion, and to the product add his poll tax, if any ; the result will be 
the whole amount of his tax. 

Note. — When a tax is to be apportioned among a large number of individuals, 
the operation is greatly facilitated by first finding the tax on $1, $2, $3, etc., to 
$9 ; then on $10, $20, $30, etc., to $90, and so on, and arranging the results as 
in the following 

TABLE. 



Prop. 


Tax. 


Prop. 


Tax. 


Prop. 


Tax. 


Prop. 


Tax. 


$1 


$.004 


$10 


$.04 


$100 


$ .40 


$1000 


$4.00 


2 


.008 


20 


.08 


200 


.80 


2000 


8. 


3 


.012 


30 


.12 


300 


1.20 


3000 


12. 


4 


.016 


40 


.16 


400 


1.60 


4000 


16. 


5 


.020 


50 


.20 


500 


2.00 


5000 


29. 


6 


.024 


60 


.24 


600 


2.40 


6000 


24. 


7 


.028 


70 


.28 


700 


2.80 


7000 


28. 


8 


.032 


80 


.32 


800 


3.20 


8000 


32. 


9 


.036 


90 


.36 


900 


3.60 


9000 


36. 



EXAMPLES FOR PRACTICE. 



1. According to the conditions of the last example, what would 
be the tax of a person whose property was valued at 2465, and 
who pays for 2 polls ? 



)g PERCENTAGE. 






OPERATION. 






From the table we find that 








The tax on $2000 




is 


$8.00 


it a a 400 




it 


1.60 


u u u go 




tt 


.24 


a a u 5 




a 


.02 


And " « ? 2 


polls 


a 


1.00 



Whole tax "-$10.86, Ans. 

2. What would A's tax be, who is assessed for $8530, and 3 
polls? Ans. $35.62. 

3. How much will C^s tax be, who is assessed for $987, and 1 
poll? Ans. $4,448. 

4. The estimated expenses of a certain town for one year are 
$6319, and the balance on hand in the public treasury is $654. 
There are 2156 polls to be assessed at $.25 each, and taxable pro- 
perty to the amount of $1864000. Besides the town tax, there 
is a county tax of 1} mills on a dollar, and a State tax of J of a 
mill on a dollar. Required the whole amount of A's tax, whose 
property is valued at $32560, and who pays for 3 polls. 

5. What does a non-resident pay, who owns property in the 
same town to the amount of $16840 ? Ans. $79.99. 

6. What sum must be assessed in order to raise a net amount 

of $5561.50, and pay the commission for collecting at 2 %. 

Note. — Since the base of the collector's commission is the sum collected, 
(446), the question is an example under Problem V of Percentage. 

7. In a certain district a school house is to be built at an ex- 
pense of $9120, to be defrayed by a tax upon property valued at 
$1536000. What shall be the rate of taxation to cover both the 
cost of the school house, and the collector's commission at 5 % ? 

8. The expenses of a school for one term were $1200 for 
salary of teachers, $57.65 for fuel, and $38.25 for incidentals ; the 
money received from the school fund was $257.75, and the remain- 
ing part of the expense was paid by a rate-bill. If the aggregate 
attendance was 9568 days, what was A's tax, who sent 4 pupils 46 
days each? Ans. $19.96+. 

9. The expense of building a public bridge was $1260.52, 



GENERAL AVERAGE. 297 

which was defrayed by a tax upon the property of the town. The 
rate of taxation was 3i mills on one dollar, and the collector's 
commission was 3 i °Jo ) what was the valuation of the property ? 

Ans. $401920. 



GENERAL AVERAGE. 

olo. General Average is a method of computing the loss to 
be sustained by the proprietors of the ship, freight, and cargo, 
respectively, when, in a case of common peril at sea, any portion 
of the property has been sacrificed or damaged for the common 
safety. 

516. The Contributory Interests are the three kinds of prop- 
erty which are taxed to cover the loss. These are, 

1st The vessel, at its value before the loss. 

2d. The freight, less i as an allowance for seamen's wages. 

3d. The cargo, including the part sacrificed, at its market value 
in the port of destination. 

Note. — In New York only £ of the freight is made contributory to the loss. 

5V7. Jettson is the portion of goods thrown overboard. 

518. The loss which is subject to general average includes, 

1st. Jettson, or property thrown overboard. 

2d. Repairs to the vessel, less i on account of the superior 
worth of the new articles furnished. 

3d. Expense of detention to which the vessel is subject in port. 

1. The ship Nelson, valued at $52000, and having on board a 
cargo worth $18000, on which the freight was $3600, threw over- 
board a portion of the goods valued at $5000, to escape wreck in 
a storm ; she then put into port, and underwent repairs amounting 
to $1200, the expenses of detention being $350. What portion 
of the loss will be sustained by each of the three contributing 
interests ? What will be paid or received by the owners of the 
ship and freight ? What by A, who owned $8000 of the cargo, 
including $3500 of the portion sacrificed, and by B, who owned 
$6000 of the cargo, including $1500 of the portion sacrificed, and 
by C, who owned $4000, or the residue of the cargo ? 



298 PERCENTAGE. 

OPERATION. 
LOSSES. CONTRIBUTORY INTERESTS. 

Jettson, $5000 Vessel, $52000 

Eepairs, less i, 800 Freight, less i, 2400 

Cost of detention, 350 Cargo, 18000 

Total,... $6150 Total, $72400 

$6150 -r- $72400 = .0849447+, rate per cent, of loss. 
$52000 X .0849447 = $4417.13, payable by vessel. 
2400 X .0849447 = 203.87, " " freight. 

18000 X .0849447 = 1529.00, " " cargo. 

$6150.00, Total contribution. 

$8000 X .0849447 = $679.56, payable by A. 
6000 X .0849447 = 509.67, " " B. 
4000 X .0849447 = 339.78, " " C. 

$4417.13 + $203.87 = $4621.00, payable by owners of vessel and freight. 
800.00+ 350.00= 1150.00, " to 
4621.00—1150.00= 3471.00, balance payable by ship owners. 
3500.00— 679.56= 2820.44, " receivable by A. 
1500.00— 509.67= 990.33, " " " B. 

Hence the following 

Rule. I. Divide the sum of the losses by the sum of the con- 
tributory interests ; the quotient will be the rate of contribution. 

II. Multiply each contributory interest by the rate ; the products 
will be the respective contributions to the loss. 

EXAMPLES FOR PRACTICE. 

1. The ship Nevada, in distress at sea, cut away her mainmast, 
and cast overboard } of her cargo, and then put into Havana to 
refit ; the repairs cost $1500, and the necessary expenses of deten- 
tion were $420. The ship was owned and sent to sea by George 
Law, and was valued at $25000; the cargo was owned by Hayden 
& Co., and consisted of 2800 barrels of flour, valued at $9 per 
barrel, upon which the freight was $4200. In the adjustment of 
the loss by general average, how much was due from Law to 
Hayden & Co.? Ans. $2629.36. 

2. A coasting vessel valued at $28000, having been disabled in 
a storm, entered port, and was refitted at an expense of $270 for 
repairs, and $120 for board of seamen, pilotage, .and dockage. 



CUSTOM HOUSE BUSINESS. 299 

Of the cargo, valued at §5000, $2400 belonged to A, $1850 to B, 
and $750 to C ; and the amount sacrificed for the ship's safety- 
was $1400 of A's property, and $170 of B's ; the gross charges 
for freight were $1500. Required the balance, payable or re- 
ceivable, by each of the parties, the loss being apportioned by 
general average. 

j f $1295 payable by ship owners; $1278 receivable by A; 
^'1 41.25 « " Cj 68.25 " " B. 

CUSTOM HOUSE BUSINESS. 

519. Duties, or Customs, are taxes levied on imported goods, 
for the support of government and the protection of home industry. 

«S20. A Custom House is an office established by government 
for the transaction of business relating to duties. 

It is lawful to introduce merchandise into a country only at 

points where custom houses are established. A seaport town 

having a custom house, is called a 'port of entry. To carry on 

foreign commerce secretly, without paying the duties imposed by 

law, is smuggling. 

Note. — Customs or duties form the principal source of revenue to the General 
Government of the United States : by increasing the price of imported goods 
they operate as an indirect tax upon consumers, instead of a general direct tax. 

521. Duties are of two kinds — Ad Valorem and Specific. 
Ad Valorem Duty is a sum computed on the cost of the goods 

in the country from which they were imported. 

Specific Duty is a sum computed on the weight or measure of 
the goods, without regard to their cost. 

522, An Invoice is a bill of goods imported, showing the 
quantity and price of each kind. 

, 523. By the New Tariff Act, approved March 2, 1857, all 
duties taken at the IL S. custom houses are ad valorem. The 
principal articles of import are classified, and a fixed rate is im- 
posed upon each list or schedule, certain articles being excepted 
and entered free. 

In collecting customs it is the design of government to tax 
only so much of the merchandise as will be available to the im- 



300 PERCENTAGE. 

porter in the market. The goods are weighed, measured, gauged, 
or inspected, in order to ascertain the actual quantity received in 
port; and an allowance is made in every case of waste, loss, or 
damage. 

524. Tare is an allowance for the weight of the package or 
covering that contains the goods. It is ascertained, if necessary, 
by actually weighing one or more of the empty boxes, casks, or 
coverings. In common articles of importation, it is sometimes 
computed at a certain per cent, previously ascertained by frequent 
trials by weighing. 

525. Leakage is an allowance on liquors imported in casks 
or barrels, and is ascertained by gauging the cask or barrel in 
which the liquor is imported. 

526. Breakage is an allowance on liquors imported in 
bottles. 

527. Gross Weight or Value is the weight or value of the 
goods before any allowance has been made. 

528. Net Weight or Value is the weight or value of the 
goods after all allowances have been deducted. 

Notes. — 1. Draft is an allowance for the waste of certain articles, and is 
made only for statistical purposes ; it does not affect the amount of duty. 

2. Long ton measure is employed in the custom houses of the United States, 
in estimating goods by the ton or hundred weight. 

The rates of this allowance are as follows : 

On 112 lb 1 lb. 

Above 112 lb. and not exceeding 224 lb., 2 lb. 

" 224 lb. " " " 336 lb., 3 lb. 

" 336 lb. " " " 1120 lb., 4 lb. 

" 1120 lb. " « " 2016 lb., 7 lb. 

" 2016 lb *... 9 lb. 

529. In all calculations where ad valorem duties are consid- 
ered, 

I. The net value of the merchandise is the worth of the net 
weight or quantity at the invoice price, allowance being made in 
cases of damage. 

II. The duty is computed at a certain legal per cent, on the 

net value of the merchandise. 

Note. — In the following examples the legal rates of duty, according to the 
New Tariff Act, are given. 



CUSTOM HOUSE BUSINESS. 301 

EXAMPLES FOR PRACTICE. 

1. What is the duty, at 24 %, on an invoice of cassimere goods 

which cost $750 ? 

Analysis. According to Prob. I, 

* 7 JHr^ian (449) > we multiply the invoice > $750 ' 

$7*>U X .^4 = %LW which is the base of the dutyj by the 

given rate, and obtain the duty, $180. 

2. The gross weight of 3 hogsheads of sugar is 1024 lb., 1016 
lb., and 1020 lb. respectively ; the invoice price of the sugar 7| 
cents, and the allowance for tare 80 lb. per hogshead ; what is the 
duty, at 24 % ? 

operation. Analyses. We first find 

1024 the gross weight of the 

1016 three boxes, from which 

1020 we subtract the tare, and 

3060, gross weight. obtain 2820 lb., the net 

80 X 3 = 240, tare. weight. We next find the 

OOOA . ," value of the net weight, 

2820, net weight. . ^, . .. . & . ' 

gr\ni at 7 J cents, the invoice 

JL price, and then compute 



$211.50, net value. the duty at 24 % on this 

.24 value, and obtain $50.76, 



$60.7600, duty. the dut 7 ™<Fired. 

3. Having paid the duty at 8 % on a quantity of Malaga 
raisins, I find that the whole cost in store, besides freight, is $378 ; 
what were the raisins invoiced at ? 

Analysis. According to Prob. 

OPERATION. jy^ (452)j we diyide the amQUntj 

$378 -v- 1.08 = $350 $ 378) by i p i us t he rate, 1.08, and 

obtain the base, or invoice, $350. 

4. A Boston jeweler orders from Lubec a quantity of watch 
movements, amounting to $2780 ; what will be the duty, at 4 % ? 

5. What will be the duty at 15 % on 1200 lb. of tapioca, in- 
voiced at 5^ cents per pound ? Ans. $9.90. 

6. What is the duty at 15 % on 54 boxes of candles, each 
weighing 1 cwt, invoiced at 8f cents per pound, allowing tare at 
3| per cent. ? 

26 



302 PERCENTAGE. 

7. A merchant imported 50 casks of port wine, each contain- 
ing originally 36 gallons, invoiced at $2.50 per gallon. He paid 
freight at $1.30 per cask, and duty at 30 %, 1| % leakage being 
allowed at the custom house, and $8.50 for cartage ; what did the 
wine cost him in store ? Arts. $5903.25. 

8. A liquor dealer receives an invoice of 120 dozen bottles of 
porter, rated at $1.25 per dozen; if 2 °/ of the bottles are found 
broken, what will be the duty at 24 % ? Ans. $35.28. 

9. The duty at 19 °f on an importation of Denmark satin was 
$614.40 ; what was the invoice of the goods ? Ans. $3260. 

10. The duty on 600 drums of figs, each containing 14 lb., 
invoiced at 5| *cents per pound, was $35.28; required, the rate 
of duty. Ans. 8 %. 

11. A merchant in New York imports from Havana 200 hhd. 
of W. I. molasses, each containing 63 gallons, invoiced at $.30 
per gallon; 150 hhd. of B. coffee sugar, each containing 500 
pounds, invoiced at $.05 per pound ; 80 boxes of lemons, invoiced 
at $2.50 per box; and 75 boxes of sweet oranges, invoiced at 
$3.00 per box. What was the whole amount of duty, estimated 
at 24 % on molasses and sugar, and at 8 % on lemons and oranges ? 

Ans. $1841.20. 

12. A merchant imported 56 casks of wine, each containing 36 
gallons net, the duty at 30 °J amounting to $907.20 + ; at what 
price per gallon was the wine invoiced ? 

13. The duty on an invoice of French lace goods at 24 %, was 
$132, an allowance of 12 <f having been made at the custom 
house for damage received since the goods were shipped ; what 
was the cost or invoice of the goods. Ans. $625. 

14. A quantity of Valencias, invoiced at $1654, cost me 
$1980.50 in store, after paying the duties and $12.24 for freight; 
what was the rate of duty ? 

15. The duty on an importation of Bay rum, after allowing 
2 % for breakage, was $823.20, and the invoice price of the rum 
was $.25 per bottle; how many dozen bottles did the importer 
receive duty at 24 %'? Ans. 1143^ doz. 



SIMPLE INTEREST. 



303 



SIMPLE INTEREST. 

530. Interest is a sum paid for the use of money. 

531. Principal is the sum for the use of which interest is 
paid. 

532. Rate per cent, per annum is the sum per cent, paid 
for the use of $100 annually. 

Note. — The rate per cent, is commonly expressed decimally as hundredths 

(442). 

5 till, Amount is the sum of the principal and interest. 

534, Simple Interest is the sum paid for the use of the 
principal only, during the whole time of the loan or credit. 

535. Legal Interest is the rate per cent, established by law. 
It varies in different States, as follows ; 



Alabama, 8 per cent. 

Arkansas, 6 " 

California, 10 " 

Connecticut, 6 " 

Delaware, 6 " 

Dist. of Columbia, . . 6 " 

Florida, 8 " 

Georgia, 7 " 

Illinois, 6 " 

Indiana, 6 u 

Iowa, 7 " 

Kentucky 6 '* 

Louisiana, 5 " 

Maine, 6 " 

Maryland, 6 " 

Massachusetts, 6 " 

Michigan, 7 " 



Minnesota, 7 per cent. 

Mississippi, 8 " " 

Missouri, 6 " " 

New Hampshire, 6 " " 

New Jersey, 6 " " 

New York, 7 " " 

North Carolina, 6 " " . 

Ohio, 6 " " 

Pennsylvania, 6 " " 

Rhode Island, 6 " " 

South Carolina, 7 " " 

Tennessee, 6 " " 

Texas, 8 " " 

United States (debts),.. 6 " " 

Vermont, 6 " " 

Virginia, 6 " " 

Wisconsin, 7 " " 



Notes. — 1. The legal rate in Canada, Nova Scotia, and Ireland is 6 per cent, 
and in England and France 5 per cent. 

2. When the rate per cent, is not specified in accounts, notes, mortgages, 
contracts, etc., the legal rate is always understood. 

3. In some States the laws allow parties to give and take higher rates, by- 
special agreement. 

-4. Book accounts bear interest after the expiration of the term of credit, and 
notes are on interest after they become due, though no mention of interest be 
made in them. 

5. If notes are to draw interest from their date, or from a given time after 
date, the fact must be so stated in the body of the notes. 

#536. Usury is illegal interest, or a greater per cent, than the 
le2:al rate. 



Note.- 
States. 



-The taking of usury is prohibited, under various penalties, in different 



304 PERCENTAGE. 

53 7. In the operations of interest there are five parts or ele- 
ments, namely : 

I. Rate per cent, per annum ; which is the fraction or decimal 
denoting how many hundredths of a number or sum of money are 
to be taken for a period of 1 year. 

II. Interest; which is the whole sum taken for^the whole period 
of time, whatever it may be. 

III. Principal ; which is the base or sum on which interest is 
computed. 

IV. Amount; which is the sum of principal and interest; and 

V. Time. 

TO COMPUTE INTEREST 
CASE I. 

538. To find the interest on any sum, at any rate 
per cent, per annum, for years and months. 

Analysis. In interest, any rate per cent, is confined to 1 year. 
Therefore, if the time be more than 1 year, the per cent, will be greater 
than the rate per cent, per annum, and if the time be less than 1 
year, the per cent, will be less than the rate per cent, per annum. 
From these facts, we deduce the following principles : 

I. If the rate per cent, per annum be multiplied by the time, 
expressed in years and fractions or decimals of a year, the product 
will be the rate for the required time. And 

II. If the principal .be multiplied by the rate for the required 
time, the product will be the required interest. Hence 

III. Interest is always the product of three factors, namely, 
rate per cent, per annum, time, and principal. 

In computing interest the three factors may be taken in any order ; 
thus, if the principal be multiplied by the rate per cent, per annum, 
the product will be the interest for 1 year ; and if the interest for 1 
year be multiplied by the time expressed in years, the result will be 
the required interest. Hence the following 

Rule. I. Multiply the principal by the rate per cent, and the 
product will be the interest for 1 year, 

II. Multiply this product by the time in years and fractions of 
a year ; the result will be the required interest 



SIMPLE INTEREST. 305 

Or, Multiply together the rate per cent per annum, time, and 
principal, in such order as is most convenient; the continued pro- 
duct will he the required interest. 

CASE II. 

539. To find the interest on any sum, for any time, 
at any rate per cent. 

The analysis of our rule is based upon the following 

Obvious Relations between Time and Interest. 
I The interest on any sum for 1 year at 1 per cent., is .01 of 
that sum, and is equal to the principal with the separatrix re- 
moved two places to the left. 

II. A month being j 1 ^ of a year, j 1 ^ of the interest on any sum 
for 1 year is the interest for 1 month. 

III. The interest on any sum for 3 days is ^ = yL = 1 of the 
interest for 1 month, and any number of days may readily be re- 
duced to tenths of a month by dividing by 3. 

IV. The interest on any sum for 1 month, multiplied by any 
given time expressed in months and tenths of a month, will pro- 
duce the required interest. 

These principles are sufficient to establish the following 
Rule. I. To find the interest for 1 yr. at 1 % : — Remove the 
separatrix in the given principal two places to the left, 

II. To find the interest for 1 mo. at 1 % : — Divide the interest 
for 1 year by 12. 

Ill To find the interest for any time at 1 % : — Multiply the 
j interest for 1 month by the given time expressed in months and 
j tenths of a month. 

IV. To find the interest at any rate % : — Multiply the interest 
i at 1 % f or the given time by the given rate. 

Contractions. After removing the separatrix in the principal two 

places to the left, the result may be regarded either as the interest 

j on the given principal for 12 months at 1 per cent., or for 1 month at 

, 12 per cent. If we regard it as for 1 month at 12 per cent., and if 

the given rate be an aliquot part of 12 per cent., the interest on the 

26* U 



306 PERCENTAGE. 

given principal for 1 month may readily be found, by taking such an 
aliquot part of the interest for 1 month as the given rate is part of 12 
per cent. Thus, 

To find the interest for 1 month at 6 pep cent., remove the separa- 
trix two places to the left, and divide by 2. 

To find it at 3 per cent., proceed as before, and divide by 4 ; at 4 
per cent., divide by 3 ; at 2 per cent., divide by 6, etc. 

SIX iPER CENT- METHOD.* 

540. By referring to 536 it will be seen that the legal rate 
of interest in 21 States is 6 per cent. This is a sufficient reason 
for introducing the following brief method into this work : 

Analysis. At 6 f per annum the interest on $1 

For 12 months is $.06. 

" 2months( T 2 2=iofl2mo.) " .01. 

" 1 month, or 30 days (^ of 12 mo.) " .00J == $.005 (^ of $.06). 

" 6 days (J of 30 da.) " .001. 

" 1 " (i of 6 da. = rir of 30 da " • 0Q °i- 
Hence we conclude that, 

1st. The interest on $1 is $.005 per month, or $.01 for every 
2 months; 

2d. The interest on $1 is $.000£ per day, or $.001 for every 6 
days. 

From these principles we deduce the 

Rule. I. To find the rate: — Call every year $.06, every 2 
months $.01, every 6 days $.001, and any less number of days 
sixths of 1 mill. 

II. To find the interest: — Multiply the principal by the rate. 

Notes. — 1. To find the interest at any other rate ^ by this method, first find 
it at 6 ft, and then increase or diminish the result by as many times itself as 
the given rate is units greater or less than 6 ft- Thus, for 7 %> add l, for 4 <jf 
subtract £, etc. 

2. The interest of $10 for 6 days, or of $1 for 60 days, is $.01. Therefore, if the 
principal be less than $10 and the time less than 6 days, or the principal less 
than $1 and the time less than 60 days, the interest will be less than $.01, and 
may be disregarded. 

3. Since the interest of $1 for 60 days is $.01, the interest of $1 for any num- 

* This method of finding the interest on $1 by inspection was first published 
in The Scholar's Arithmetic, by Daniel Adams, M. D., in 1801, and from its 
simplicity it has come into very general use. 



SIMPLE INTEREST. 307 

ber of days is as many cents as 60 is contained times in the number of days. 
Therefore, if any principal be multiplied by the number of days in any given 
number of months and days, and the product divided by 60, the result will be 
the interest in cents. That is, Multiply the principal by the number of days, 
divide the product by 60, and point off two decimal places in the quotient. The 
result will be the interest in the same denomination as the principal. 

EXAMPLES FOR PRACTICE. 

What is tne interest on the following sums for the times given, 
at 6 per cent. ? 

1. $325 for 3 years. Ans. $58.50. 

2. $1600 tor 1 yr. 3 mo. Ans. $120. 

3. $36.84 for 5 mo. 

4. $35.14 for 2 yr. 9 mo. 15 da. 

5. $217.15 for 3 yr. 10 mo. 1 da. Ans. $49.98. 

6. $721.53 for 4 yr. 1 mo. 18 da. 

7. $15,125 for 15 mo. 17 da. Ans. $1.17+. 
On the following at 7 per cent. ? 

8. $2000 for 5 yr. 6 mo. 

9. $1436.59 for 2 yr. 5 mo. 18 da. Ans. $248,051 + . 

10. $224.14 for 8 mo. 13 da. Ans. $11,026. 

11. $100.25 for 63 da. Ans. $1,227 + . 

12. $600 for 24 da. 

13. $520 for 5 yr. 11 mo. 29 da. Ans. $218,298. 

14. $710.01 for 3 yr. 11 mo. 8 da. 
On the following at 5 per cent. ? 

15. $48,255 for 5 yr. 

16. $750 for 1 yr. 3 mo. 

17. $647,654 for 4 yr. 10 mo. 20 da. Ans. $158,315 

18. $12850 for 90 da. - 

19. $2500 for * mo. 20 da. Ans. $79.86. 

20. $850.25 for 8 mo. 

21. $48.25 for 1 yr. 2 mo. 17 da. Ans. $2,928 + . 
On the following at 8 per cent. ? 

22. $2964.12 for 11 mo. Ans. $217,367. 

23. $725.50 for 150 da. 

24. $360 for 2 yr 6 mo. 12 da. 

25. $600 for 3 yr. 2 mo. 17 da. Ans. $154.2661. 



308 PERCENTAGE. 

26. $1700 for 28 da. Ans. $10.57. 
On the following at 10 per cent. ? 

27. $3045.20 for 7 mo. 15 da. Ans. $190.32 + . 

28. $1247.375 for 2 yr. 26 da. Ans. $258.48+. 

29. $2450 for 60 da. 

30. $375,875 for 3 mo. 22 da. 
81. $5000 for 10 da. 

32. $127.65 for 1 yr. 11 mo. 3 da. Ans. $24.48—. 

23 What Is the interest of $155.49 for 3 mo., at 6i per cent. ? 

34. What is the interest of $970.99 for 6 mo., at 5 J per cent. ? 

35. What is the amount of $350.50 for 2 yr. 10 mo., at 7 per 
cent.? Ans. $420.01+. 

36. What is the interest of $95,008 for 3 mo. 24 da., at 4} per 
cent.? Ans. $1,353+. 

37. What is the amount of $145.20 for 1 yr. 9 mo. 27 da., at 
12£ per cent.? Ans. $178.32375. 

38. What is the amount of $215.34 for 4 yr. 6 mo., at 3 J per 
cent.? Ans. $249,256+. 

39. What is the amount of $5000 for 20 da., at 7 per cent. ? 

40. What is the amount of $16941.20 for 1 yr. 7 mo. 28 da., 
at 4| per cent. ? Ans. $18277.91. 

41 If $1756.75 be placed at interest June 29, 1860, what 
amount will be due Feb. 12, 1863, at 7 % ? 

42. If a loan of $3155.49 be made Aug. 15, 1858, at 6 per 
cent., what amount will be due May 1, 1866, no interest having 
been paid? 

43. How much is the interest on a note dated March 1, 1859, 
and payable July 16, 1861, at 7 % t 

44. A person borrows $3754.45, being the property of a minor 
who is 15 yr. 3 mo. 20 da. old. He retains it until the owner is 
21 years old. How much money will then be due at 6 % simple 
interest? Ans. $5037.22. 

45. If a person borrow $7500 m Boston and lend it in Wis- 
consin, how much does he gain in a year ? 

46. A man sold a piece of property for $11320; the terms were 
$3200 in cash on delivery, $3500 in 6 mo., $2500 in 10 mo., and 



SIMPLE INTEREST. 309 

the remainder in 1 yr. 3 mo. ; with 7 % interest; what was the 
whole amount paid t - Ans. $11788.41*. 

47. May 10, 1859, I borrowed $6840, with which I purchased 
flour at $5.70 a barrel. June 21, 1860, 1 sold the flour for $6.62 J 
a barrel, cash. How much did I gain by the transaction, interest 
being reckoned at 6 % ? 

48. If a man borrow $15000 in New York, and lend it in 
Ohio, how much will he gain in 1 yr. and a half, reckoning 360 
days to the year in the former transaction, and 365 days in the 
latter ? 

49. Hubbard & Northrop bought bills of dry goods of Bowen, 
McNamee & Co., New York, as follows, viz. : July 15, 1860, 
$1250; Oct. 4, 1860, $3540.84; Dec. 1, 1860, $575; and Jan. 
24, 1861, $816.90. They bought on time, paying legal interest; 
how much was the whole amount of their indebtedness, March 1, 
1861? 

50. A broker allows 6 per cent, per annum on all moneys de- 
posited with him. If on an average he lend out every $100 re- 
ceived on deposit 11 times during the year, for 83 days each 
time at 2 % a month, how much does he gain by interest on 
$1000? Ans. $182. 

'51. A man, engaged in business with a capital of $21840, is 
making 12 J per cent, per annum on his capital; but on account of 
' ill health he quits his business, and loans his money at 7f %. 
! How much does he lose in 2 yr. 5 mo. 10 da. by the change ? 

Ans. $2535.861. 

52. A speculator wishing to purchase a tract of land containing 
; 450 acres at $27.50 an acre, borrows ths money at 5 i per cent. 
j At the end of 4 yr. 11 mo. 20 da he sells § of the land at $34 

an acre, and the remainder at $32.55 an acre. How much does 
he gain by the transaction ? 

53. Bought 4500 bushels of wheat at $1.12} a bushel, payable 
jin 4 months; I immediately realized for it $1.06 a bushel, cash, 

and put the money at interest at 10 per cent. At the end of 6 
months I paid for the wheat; did I gain or lose by the transac- 
tion, and how much ? 



310 PERCENTAGE. 

PARTIAL PAYMENTS OR INDORSEMENTS. 

541. A Partial Payment is payment in part of a note, bond, 
or other obligation. 

542. An Indorsement is an acknowledgment written on the 
back of an obligation, stating the time and amount of a partial 
payment made on the obligation. 

543. To secure uniformity in the method of computing in- 
terest where partial payments have been made, the Supreme Court 
of the United States has decided that, 

I. " The rule for casting interest when partial payments have 
been made, is to apply the payment, in the first place, to the dis- 
charge of the interest then due. 

II. " If the payment exceeds the interest the surplus goes to- 
wards discharging the principal, and the subsequent interest is to 
be computed on the balance of the principal remaining due, 

IIL '< If the payment be less than the interest the surplus of 
the interest must be taken to augment the principal, but the inte- 
rest continues on the former principal until the period when the 
payments, taken together, exceed the interest due, and then the 
surplus is to be applied towards discharging the principal, and the 
interest is to be computed on the balance as aforesaid/' — Decision 
of Chancellor Kent 

This decision has been adopted by nearly all the States of the 
Union, the only prominent exceptions being Connecticut, Ver- 
mont, and New Hampshire. We therefore present the method 
prescribed by this decision as the 

United States Rule. 

I. Find the amount of the given principal to the timz of the 
first payment, and if this payment exceed the interest then due, 
subtract it from the amount obtained, and treat the remainder as a 
new principal. 

II. But if the interest be greater than any payment, compute the 
interest on the same principal to a time when the sum of the pay- 
ments shall equal or exceed tJie interest due, and subtract the sum 



PARTIAL PAYMENTS. 3H 

of the payments from the amount of the principal ; the remainder 
wiUform a new principal, with which proceed as before. 

EXAMPLES FOR PRACTICE. 

$1000- Buffalo, N Y., May 15, 1856. 

1. Two years after date I promise to pay to David Hudson, or 
order, one thousand dollars, with interest, for value received. 

Henry Burr. 
On this note were indorsed the following payments : 

Sept. 20, 1857, received, $150.60 

Oct. 25,1859, " 200.90 

July 11, 1861, " 75.20 

Sept. 20,1863, " 112.11 

Dec. 5,1863, " 105. 

What remained due May 20, 1864 ? 

OPERATION. 

i 

Principal on interest from May 15, 1856, $1000 

Interest to Sept. 20, 1857, 1 yr. 4 mo-. 5 da., 94.31 

Amount, $1094.31 

1st Payment, Sept 20, 1857, 150.60 

Remainder for a new principal, $943.71 

Interest from 1st paym't to Oct. 25, 1859, 2 yr. 1 mo. 5 da., 138.54 

Amount, $1082.25 

2d Payment, Oct. 25, 1859, 200.90 

Eemainder for a new principal, $881.35 

I Interest from 2d paym't, Sept. 20, 1862, 2 yr. 10 mo. 25 da., 179.09 

Amount, $1060.44 

3d Payment, less than interest due, ... $75.20 

4th " 112.11 

187,31 

i Remainder for a new principal, $873.13 

Interest from 4th paym't to Dec. 5, 1863, 1 yr. 2 mo, 15 da., 73.86 

Amount, $946 99 

5th Payment, Dec. 5, 1863, 105.00 

Remainder for a new principal, $841,99 

j Interest from 5th paym't to May 20, 1864, 5 mo. 15 da., . . 27.01 

Balance due May 20, 1864, $869.00 



322 PERCENTAGE. 



$1200. Richmond, Va., Oct. 15, 1859. 

2. One year after date we promise to pay James Peterson, or 
order, twelve hundred dollars, for value received, with interest. 

Wilder & Son. 

Indorsed as follows: Oct. 15, 1860, $1000; April 15, 1861, 
$200. How much remained due Oct. 15, 1861 ? Am. $82.56. 



$850 T ^. Boston, June 10, 1855. 

3. Eighteen months after date I promise to pay Crosby, Nich- 
ols & Co., or order, eight hundred fifty and y 7 ^ dollars, with 
interest, for value received. O. L. Sanborn. 

Indorsed as follows: March 4, 1856, $210.93; July 9, 1857, 
$140; Feb. 20, 1858, $178; May 5, 1859, $154.30; Jan. 17, 
" 1860, $259 45. How much was due Oct. 24, 1861 ? 



$384 T 9 o 5 o - Savannah, Ga., Sept. 4, 1860. 

4. Six months after date I promise to pay John Rogers, or 
order, three hundred eighty-four and T 9 ^j dollars, for value re- 
ceived, with interest. Wm. Jenkins. 

This note was settled Jan. 1, 1862, one payment of $126.50 
having been made Oct. 20, 1861 ; how much was due at the time 
of settlement ? 



$3475. New Orleans, March 6, 1857. 

5 On demand we promise to pay Evans & Hart, or order, three 
thousand four hundred seventy-five dollars, for value received, with 
interest. Davis & Brother. 

Indorsed as follows- June 1, 1857, $1247.60; Sept 10, 1857, 
$1400. How much was due Jan. 31, 1858 ? 

6. A gentleman gave a mortgage on his estate for $9750, dated 
April 1, 1860, to be paid in 5 years, with annual interest after 9 
months on all unpaid balances, at 10 per cent. Six months from 
date he paid $846.50; Oct. 20, 1862, $2500 ; July 3, 1863, $1500; 
Jan. 1, 1864, $500; how much was due at the expiration of the 
given time? 



PARTIAL PAYMENTS. 3^3 



$ 5Q0 - Philadelphia, Feb. 1, 1861. 

7. For value received, I promise to pay J. B. Lippincott & Co., 
or order, five hundred dollars three months after date, with interest. 

James Monroe. 

Indorsed as follows: May 1, 1861, $40; Nov. 14, 1861, $8; 
April 1, 1862, $12 j May 1, 1862, $30. How much was due 
Sept. 16,1862? Arts. $455.57 

544. Connecticut Rule. 

I. Payments made one year or more from the time the interest 
commenced, or from another payment, and payments less than the 
interest due, are treated according to the United States rule, 

II. Payments exceeding the interest due and made within one 
year from the time interest commenced, or from a former payment, 
shall draw interest for the balance of the year, provided the interval 
does not extend beyond the settlement, and the amount must be sub- 
tracted from the amount of the principal for one year ; the re- 
mainder will be the new principal, 

III. If the year extend beyond the settlement, then find the 
amount of the payment to the day of settlement, and subtract it 
from the amount of the principal to that day ; the remainder will 
be the sum due, 

5^k5. A note containing a promise to pay interest annually 
is not considered in law a contract for any thing more than simple 
interest on the principal. For partial payments on such notes 
the following is the 

Vermont Rule. 

I. Find the amount of the principal from the time interest com- 
menced to the time of settlement, 

II. Find the amount of each payment from the time it icas made 
to the time of settlement. 

III. Subtract the sum of the amounts of the payments from the 
amount of the principal ; the remainder will be the sum due. 

Note.- — This rule is in quite extensive use among merchants and others. 

27 



314 



PERCENTAGE. 



•546. In New Hampshire interest is allowed on the annual 
interest if not paid when due, in the nature of damages for its 
detention ; and if payments are made before one year's interest 
has accrued, interest must be allowed on such payments for the 
balance of the year. Hence the following 

New Hampshire Rule. 

I. Find the amount of the principal for one year, and deduct 
from it the amount of each payment of that year, from the time 
it was made up to the end of the year ; the remainder will be a 
new principal, with which proceed as before. 

II. If the settlement occur less than a year from the last annual 
term of interest, make the last term^f interest a part of a year, 
accordingly. 

EXAMPLES FOR PRACTICE. 



$1000 - New Haven, Conn., Feb. 1, 1856. 

1. Two years after date, for value received, I promise to pay to 
Peck & Bliss, or order, one thousand dollars with interest. 

John Cornwall. 
Indorsed as follows: April 1, 1857, $80; Aug. 1, 1857, $30; 
Oct. 1, 1858, $10 ; Dec. 1, 1858, $600 ; May 1, 1859, $200. How 
much was due Oct. 1, 1859? Ans. $263.93. 

ft 2000 - Burlington, Vt., May 10, 1858. 

2. For value received, I promise to pay David Camp, or order, 
two thousand dollars, on demand, with interest annually. 

Kichard Thomas. 
On this note were indorsed the following payments : March 10, 
1859, $800; May 10, 1860, $400; Sept. 10, 1861, $300. How 
much was due Jan. 10, 1863 ? 

3. How much would be due on the above note, computing by 
the Connecticut rule ? Ans. $831,737. 

4. How much, computing by the New Hampshire rule? By 
the United States rule ? * ( N. H. rule, $833.21 ; 

1 U. S. " $831.90. 



SAVINGS BANKS ACCOUNTS. 3X5 

SAVINGS BANKS ACCOUNTS. 

547. Savings Banks are institutions intended to receive in 
trust or on deposit, small sums of money, generally the surplus 
earnings of laborers, and to return the same at a moderate interest 
at a future time. 

548. It is the custom of all savings banks to add to each 
depositor's account, at the end of a certain fixed term, the interest 
due on his deposits according to some general regulation for allow- 
ing interest. The interest term with some savings banks is 6 
months, with some 3 months, and with some 1 month. 

o40. A savings bank furnishes each depositor with a book, 
in which is recorded from time to time the sums deposited and 
the sums drawn out. The Dr. side of such an account shows the 
deposits, and the Cr. side the depositor's checks or drafts. In the 
settlement, interest is never allowed on any sum which has not 
been on deposit for a full interest term. Hence, to find the 
amount due on any depositor's account, we have the following 

Rule. At the end of each term, add to the balance of the 
account one term's interest on the smallest balance on deposit at any 
one time during that term ; the final balance thus obtained will be 
the sum due. 

Notes. — 1. It will be seen that by this rule no interest is allowed for 
money on deposit during a partial term, whether the period be the first or the 
last part of tbe term. 

2. An exception to this general rule occurs in the practice of some of the 
savings banks of New York city. In these, the interest term is 6 months, and 
the depositor is allowed not only the full term's interest on the smallest balance, 
but a half term's interest on any deposit, or portion of a deposit made during 
the first 3 months of the term, and not drawn out during any subsequent part <>f 
the term. 

EXAMPLES FOR PRACTICE. 

1. What will be due April 20, 1860, on the following account, 
interest being allowed quarterly at 6 per cent, per annum, the 
terms commencing Jan. 1, April 1, July 1, and Oct. 1 ? 

Dr. Savings Bank in account with James Taylor. Cr. 

1858, Jan. 12, .$75 1858, March 5, $30 

" May 10, ., 150 " Aug. 16, 50 

H Sept. 1, 20 " Dec. 1, 48 

1859, Feb. 16, 130 



316 



PERCENTAGE. 



OPERATION. 

Deposit, Jan. 12, 1858, $75 

Draft, March 5, " _ J30 

Balance, Apr. 1, 1860, $4.5 

Deposit, May 10, 1858, 150 

Int. on $45, for 3 mo .-. .68 

Balance, July 1, 1860, $195.68 

Draft, Aug. 16, 1858, 50 

Least balance during the current term,.., $145.68 

Deposit, Sept. 1, 1858, 20.00 

Ink on $145.68, for 3 mo. 2.19 

Balance, Oct. 1, 1858, $167.87 

Draft, Dec. 1, 1858, 48 

Least balance during the current term, 119.87 

Int. on $119.87, for 3 mo 1.80 

Balance, Jan. 1, 1860, ,$121.67 

Deposit, Feb. 16,1860, 130.00 

Int. on $121.67, for 3 mo 1.83 

Bal. due after Apr. 1, 1860, $253.50 Ans. 

Note. — In the following examples the terms commence with the year, or on 
Jan. 1. 

2. Allowing interest monthly at 6 % per annum, what sum 
will be due Sept. 1, I860, on the book of a savings bank having 
the following entries ? 

Bay State Savings Institution, in account with Jane Ladd. 



Dr. 



Or. 



1860. 










1860. 




Jan. 


3 


To cash, 


5 


75 


Jan. 


28 


it 


8 


t( (( 


13 


45 


Feb. 


7 


n 


20 


ti a 


7 


60 


March 


20 


Feb. 


20 


" check, 


16 


45 


April 


11 


« 


27 


" cash, 


8 


40 


June 


3 


March 


6 


(t check, 


14 


65 


ei 


12 


" 


29 


" cash, 


7 


98 


u 


20 


April 


25 


« n 


3 


49 


Aug. 


17 


May 
u 


7 
30 


" draft, 


26 
45 


50 
79 






July 


28 


" cash, 


15 


68 






Aug. 


3 


" check, 


18 


45 








26 


u cash, 


4 


50 







By check, 



" draft, 
" check, 



k, 


5 


00 




8 


48 




10 


00 




12 


76 




3 


96 




10 


48 




17 


48 


k, 


5 


64 



Ans. $106.70. 
3. Interest at 7 %, allowed quarterly, how much was due April 
4, 1860, on the following savings bank account ? 



COMPOUND INTEREST. 



317 



Dr. 



Detroit Savings Institution, in account with E. L. Selden. 



Cr. 



1859. 










1859. 










Jan. 


1 


To cash, 


. 47 


50 


May 


12 


By check, 


50 


36 


March 


12 


u ii 


124 


36 


Oct. 


3 


u a 


25 


78 


June 


20 


ii (( 


130 


56 


Not. 


16 


a a 


36 


48 


Aug. 


3 


a « 


68 


75 


Dec. 


28 


« a 


12 


50 


1860. 










i 










Jan. 


25 


a a 


160 


80 


1 











Ans. $421.82. 
3. How much was due Jan. 1, I860, on the following account, 
allowing interest semi-annually, at 6 % per annum ? 

Irvings Savings Institution, in account with James Taylor. 



Dr. 



Cr. 



1858. 










1858. 




June 


4 


To cash, 


175 




Sept. 


14 


Nov. 


1 


u u 


150 




1859. 




1859. 










July 


25 


Feb. 


24 


« draft, 


200 




Dec. 


3 


Sept. 


10 


" check, 


56 









By check, 



65 



120 
80 



4. Interest at 5 %, 
due, Jan. 1, 1860, on 
New York, having the 



Ans. $339.09. 
allowed according to Note 2, how much was 
the book of a savings bank in the city of 
following entries? 



Dr. 



Sixpenny Savings Bank, in account with William Gallup. 



Cr. 



1858. 










1858. 










June 


1 


To check, 


36 


50 


Sept. 


16 


By check, 


36 


16 


March 


17 


a tc 


25 


88 


1859. 










Aug. 


1 


" cash, 


84 


72 


Jan. 


27 


tt a 


13 


48 


1859. 










March 


1 


u u 


17 


50 


June 


11 


" draft, 


50 


00 












Nov. 


16 


" cash, 


40 


78 













Ans. $179.10. 



COMPOUND INTEREST. 

550. Compound Interest is interest on both principal and 
interest, when the interest is not paid when due. 

Note. — The simple interest maybe added to the principal annually, semi- 
annually, or quarterly, as the parties may agree ; but the taking of compound 
interest is not legal. 

1 . What is the compound interest of $640 for 4 years, at 5 
per cent. ? 

27* 



318 PERCENTAGE. 

OPERATION. 

$640 Principal for 1st year, 

$640 x 1.05 = $672 " " 2d " 

$672 x 1.05 = $705.60 M « 3d « 

$705.60 x 1.05 = $740.88 " " 4th » 

$740.88 x 1.05 = $777,924 Amount " 4 years, 
640. Given principal, 

$137,924 Compound interest. 
This illustration is sufficient to establish the following 
Rule. I. Find the amount of the given principal at the given 
rate for one year , and make it the principal for the second year. 

II. Find the amount of this new principal, and make it the 
principal for the third year, and so continue to do for the given 
number of years. 

III. Subtract the given principal from the last amount ; the re- 
mainder will be the compound interest 

Notes. — 1. When the interest is payable semi-annually or quarterly, find the 
amount of the given principal for the first interval, and make it the principal 
for the second interval, proceeding in all respects as when the interest is payable 
yearly. 

2. When the time contains years, months, and days, find the amount for the 
years, upon which compute the interest for the months and days, and add it to 
the last amount, before subtracting. 

EXAMPLES FOR PRACTICE. 

1. What is the compound interest of $750 for 4 years at 6 per 
cent. ? Ans. $196,857. 

2. What will $250 amount to in 3 years at 7 per cent, compound 
interest? Ans. $306.26. 

3. At 7 per cent, interest, compounded semi-annually, what 
debt will $1475.50 discharge in 2 } years? Ans. $1752.43. 

4. Find the compound interest of $376 for 3 yr. 8 mo. 15 da., 
at 6 per cent, per annum. Ans. $90.84. 

55% A more expeditious method of computing compound 
interest than the preceding is by the use of the compound interest 
tables on the following page. 



COMPOUND INTEREST. 



319 



TABLE, 

Showing the amount of $1, or £1, at 2£, 3, 3J, 4, 5, 6, 7, and 8 per 

cent., compound interest, for any number of years from 1 to 40. 



Years 

1 


2%perct. 


3 percent. 


3%perct. 


4 per cent. 


5 per cent. 


6 per cent. 


7 per cent. 


8 per cent. 


1.025000 


1.030000 


1.035000 


1.040000 


1.050000 


1.060000 


1.070000 


1.080000 


2 


1.050625 


1.060900 


1071225 


1.081600 


1.102500 


1.123600 


1.144900 


1.166400 


3 


1.076891 


1.092727 


1.108718 


1.124864 


1.157625 


1.191016 


1.225043 


1.259712 


4 


1.103813 


1.125509 


1.147523 


1.169859 


1.215506 


1.262477 


1.310796 


1.360489 


5 


1.131408 


1.159274 


1.187686 


1.216653 


1.276282 


1.338226 


1.402552 


1.469328 


6 


1.159693 


1.194052 


1.229255 


1.265319 


1.340096 


1.418519 


1.500730 


1.586874 


7 


1.188686 


1.229874 


1.272279 


1.315932 


1.407100 


1.503630 


1.605782 


1.713824 


8 


1.218403 


1.266770 


1.316809 


1.368569 


1.477455 


1.593848 


1.718186 


1.850930 


9 


1.248863 


1.304773 


1.362897 


1.423312 


1.551328 


1.689479 


1.838459 


1.999005 


\) 


1.280085 


1.343916 


1.410599 


1.480244 


1.628885 


1.790848 


1.967151 


2.158925 


11 


1.312087 


1.384234 


1.459970 


1.539454 


1.710339 


1.898299 


2.104852 


2.331639 


12 


1.344889 


1.425761 


1.511069 


1.601032 


1.795856 


2.012197 


2.252192 


2.518170 


13 


1.378511 


1.468534 


1.563956 


1.665074 


1.885649 


2.132928 


2.409845 


2.719624 


14 


1.412974 


1.512590 


1.618695 


1.731676 


1.979932 


2.260904 


2.578534 


2.937194 


15 


1.448298 


1.557967 


1.675349 


1.800944 


2.078928 


2.396558 


2.759032 


3.172169 


16 


1.484506 


1.604706 


1.733986 


1.872981 


2.182875 


2.540352 


2.952164 


3.425943 


17 


1.521618 


1.652848 


1.794676 


1.947901 


2.292018 


2.692773 


3.158815 


3.700018 


18 


1.559659 


1.702433 


1.857489 


2.025817 


2.406619 


2.854339 


3.379932 


3.996020 


19 


1.598650 


1.753506 


1.922501 


2.106849 


2.526950 


3.025600 


3.616528 


4.315701 


20 


1.638616 


1.806111 


1.989789 


2.191123 


2.653298 


3.207136 


3.869685 


4.660957 


21 


1.679582 


1.860295 


2.059431 


2.278768 


2.785963 


3.399564 


4.140562 


5.033834 


22 


1.721571 


1.916103 


2.131512 


2.369919 


2.925261 


3.603537 


4.430402 


5.436540 


23 


1.764611 


1.973587 


2.206114 


2.464716 


3.071524 


3.819750 


4.740530 


5.871464 


24 


1.808726 


2.032794 


2.283328 


2.563304 


3.225100 


4.048935 


5.072367 


6.341181 


25 


1.853944 


2.093778 


2.363245 


2.665836 


3.386355 


4.291871 


5.427433 


6.848475 


26 


1.900293 


2.156591 


2.445959 


2.772470 


3.555673 


4.549383 


5.807353 


7.396353 


27 


1.947800 


2.221289 


2.531567 


2.883369 


3.733456 


4.822346 


6.213868 


7.988062 


28 


1.996495 


2.287928 


2.620172 


2.998703 


3.920129 


5.111687 


6.648^38 


8.627106 


29 


2.046407 


2.356566 


2.711878 


3.118651 


4.116136 


5.418388 


7.114257 


9.317275 


30 


2.097568 


2.427262 


2.806794 


3.243398 


4.321942 


5.743491 


7.612255 


10.062657 


31 


2.150007 


2.500080 


2.905031 


3.373133 


4.538040 


6.088101 


8.145113 


10.867669 


32 


2.203757 


2.575083 


3.006708 


3.508059 


4.764942 


6.453387 


8.715271 


11.737083 


33 


2.258851 


2.652335 


3.111942 


3.648381 


5.003189 


6.840590 


9.325340 


12.676050 


34 


2.315322 


2.731905 


3.220860 


3.794316 


5.253348 


7.251025 


9.978114 


13.690134 


35 


2.373205 


2.813862 


3.333590 


3.946089 


5.516015 


7.686087 


10.676582 


14.785344 


36 


2.432535 


2.898278 


3.450266 


4.103933 


5.791816 


8.147252 


11.423942 


15.968172 


37 


2.493349 


2.985227 


3.571025 


4.268090 


6.081407 


8.636087 


12.223618 


17.245626 


38 


2.555682 


3.074783 


3.696011 


4.438813 


6.385477 


9.154252 


13.079271 


18.625276 


39 


2.619574 


3.167027 


3.825372 


4.616366 


6.704751 


9.703508 


13.994820 


20.115298 


40 


2.685064 


3.262038 


3.959260 


4.801021 


7.039989 


10.285718 


14.974458 


21.724522 



320 PERCENTAGE. 

EXAMPLES FOR PRACTICE. 

1. What is the amount of $300 for 4 years at 6 per cent, com- 
pound interest,payable semi-annually ? 

operation. Analysis. The amount of $1 at 6 per cent., 

$1 26677 compound interest payable semi-annually, is 

300 the same as the amount of $1 at 3 per cent., 

<fl»oon nqiflfl compound interest payable annually. We 

therefore take, from the table, the amount of 

$1 for 8 years at 3 per cent., and multiply this amount by the given 

principal. 

2. What is the amount of $536.75 for 12 yr. at 8 per cent, com- 
pound interest ? Arts. $1351.63. 

3. What sum placed at simple interest for 2 yr. 9 mo. 12 da., 
at 7 per cent., will amount to the same as $1275, placed at com- 
pound interest for the same time and at the same rate, payable 
semi-annually? Ans. $1292.52. 

4. At 8 per cent, interest compounded quarterly, how much 
will $1840 amount to in 1 yr. 10 mo. 20 da. ? Ans. $2137.05. 

5. A father at his death left $15000 for the benefit of his only 
son, who was 12 yr. 7 mo. 12 da. old when the money was de- 
posited ; the same was to be paid to him when he should be 21 
years of age, together with 7 per cent, interest compounded semi- 
annually. How much was the amount paid him ? 

6. What sum of money will amount to $2902.263 in 20 years, 
at 7 % compound interest ? Ans. $750. 

PROBLEMS IN INTEREST. 
PROBLEM I. 

552. Given, the time, rate per cent., and interest, to 
find the principal. 

1. What sum of money will gain $87.42 in 4 years, at 6 per 
cent. ? 

operation. Analysis. Since $.24 

$.24, interest of $1 for 4 years. is the int erest of $1 for 4 
$87.42 -r- .24 = $364.25, Ans. years at 6 per cent., $87.42 

must be the interest of as 



PROBLEMS IN INTEREST. 



321 



many dollars, for the same time and at the same rate, as $.24 is con- 
tained times in $87.42. Dividing, we obtain $364.25, the required 
principal. Hence the 

Rule. Divide the given interest hy the interest of $1 for the 
given time at the given rate. 

EXAMPLES FOR PRACTICE. 

1. What sum of money, invested at 6J per cent., will produce 
$279,825 in 1 yr. 6 mo.? Ans. $2870. 

2. What sum will produce $63.75 interest in 6 mo. 24 da. at 
7 J per cent. ? 

3. What sum will produce $12} interest in 10 days at 10 per 
cent.? Ans. $4500. 

4. What sum must be invested in real estate paying 12} per 
cent, profit in rents, to give an income of $3125 ? 

5. What is the value of a house and lot that pays a profit of 9} 
per cent, by renting it at $30 per month ? 

6. What sum of money, put at interest 6 yr. 5 mo. 11 da,, at 
7 per cent., will gain $3159.14 ? Ans. $7000. 

7. What sum of money will produce $69.67 in 2 yr. 9 mo. at 
6 °fc compound interest ? Ans. $400. 

8. What principal at 6 % compound interest will produce 
$124.1624 in 1 vr. 6 mo. 15 da. ? Ans. $540.50. 

PROBLEM II. 

553. Given, the time, rate per cent., and amount, to 

find the principal. 

1. What sum of money in 2 years 6 months, at 7 per cent., 

will amount to $136,535 ? 

operation. Analysis. Since 

$1,175, amount of $1 for 2 yr. 6 mo. $1,175 is the amount 

$136,535 ~ 1.175 = $116.20, Ans. of $\ for 2 /f rs 6 

months, at 7 per 

cent., $136,535 must be the amount of as many dollars, for the same 

time and at the same rate, as $1,175 is contained times in $136,535. 

Dividing, we obtain $116.20, the required principal. Hence the 

V 



322 PERCENTAGE. 

Rule. Divide the given amount by the amount of $1 for the 
given time at the given rate. 

EXAMPLES FOR PRACTICE. 

1. What principal in 2 yr. 3 mo. 10 da., at 5 per cent., will 
amount to $1893 61 ? Ans. $1700. 

2. A note which had run 3 yr. 5 mo. 12 da. amounted to 
$681,448, at 6 per cent. ; how much was the face of the note ? 

3. What sum put at interest at 3J per cent., for 10 yr. 2 mo., 
will amount to $15660 ? 

4. What is the interest of that sum for 2 yr. 8 mo. 29 da., at 7 
per cent., which at the same time and rate, will amount to 
$1568.97? Ans. $215.87. 

5. What is the interest of that sum for 243 days at 8 per cent., 
which at the same time and rate, will amount to $11119.70 ? 

6. What principal in 4 years at 6 per cent, compound interest, 
will amount to $8644.62? Ans. $6847.36. 

7. What sum put at compound interest will amount to $26772.96, 
due in 10 yr. 5 mo., at 6 per cent, compound interest? 

Ans. $14590.70. 

PROBLEM III. 

554:. Given, the principal, time, and interest, to find 
the rate per cent. 

1. I received $315 for 3 years' interest on a mortgage of 
$1500 ; what was the rate per cent. ? 

operation. Analysis. Since 

$15.00 $45 is the interest on 

3 the mortgage for 3 

years at 1 per cent., 



.00, int. for 3 yr. at 1 %. $315 mugt be ^ /; 

$315 -T- $45 = 7 %, Ans. terest on the mortgage 

for the same time, at 
as many times 1 per cent, as $45 is contained times in $315. Divid- 
ing, and we obtain 7, the required rate per cent. Hence the 

Rule. Divide the given interest by the interest on the principal 
for the given time at 1 per cent. 



PROBLEMS IN INTEREST. 



323 



EXAMPLES FOR PRACTICE. 

1. If I loan $750 at simple interest, and at the end of 1 yr. 3 
mo. receive $796.87?, what is the rate per cent. ? Ans. 5. 

2 If I pay $10.54 for the use of $1700, 28 days, what is the 
rate of interest ? Ans. 8 per cent. 

3. Borrowed $600, and at the end of 9 yr. 6 mo. returned 
$856.50 ; what was the rate per cent. ? 

4. A man invests 17266.28, which gives him an annual income 
of $744,793 ; what rate of interest does he receive ? 

5. If C buys stock at 30 per cent, discount, and every*6 months 
receives a dividend of 4 per cent., what annual rate of interest 
does he receive ? Ans. 11| per cent. 

6. At what rate per annum of simple interest will any sum of 
money double itself in 4, 6, 8, and 10 years, respectively ? 

7. At what rate per annum of simple interest will any sum 
triple itself in 2, 5, 7, 12, and 20 years, respectively ? 

8. A house that rents for $760.50 per annum, cost $7800 : what 
% does it pay on the investment? Ans. 9f per cent. 

9. I invest $35680 in a business that pays me a profit of $223 a 
month ; what annual rate of interest do I receive ? Ans. 7 i %. 

PROBLEM IV, 

555. Given, the principal, interest, and rate, to find 

the time. 

1. In what time will $924 gain $151,536, at 6 per cent? 

™^ imT ™ Analysis. Since 

operation. - •'.; . . 

&Q24 $55.44 is the interest 

06 of $924 for 1 year at 

tSS, »t of |M0 ft, 1 y , at « %. i£ ZlflSZ 
$151,536 ~- $55.44 = 2.73 of the same sum> at 

2.73 yr.= 2 yr. 8 mo. 24 da., Ans. t h e same rate per 

cent., for as many 
years as $55.44 is contained times in $151,536, which is 2.73 times. 
Reducing the mixed decimal to its equivalent compound number, and 
we have 2 years 8 months 24 days, the required time. Hence the 



324 PERCENTAGE. 

Rule. Divide the given interest by the interest on the principal 
for 1 year ; the quotient will be the required time in years and 
decimals. 

EXAMPLES FOR PRACTICE. 

1. In what time will $273.51 amount to $312,864, at 7 per 
cent. ? Ans. 2 yr. 20 da. 

2. How long must $650.82 be on interest to amount to $761.44, 
at 5 per cent. ? Ans. 3 yr. 4 mo. 24 da. 

3. How long will it take any sum of money to double itself by 
simple interest at 3, 4 J, 6, 7, and 10 per cent. ? How long to 
quadruple itself? ^ f To double itself at 3 % 33 J vr. 

US ' ( To quadruple itself at 3 %. 100 yr. 

4. In what time will $9750 produce $780 interest, at 2 per 
cent, a month ? 

5. How long must I keep on deposit $439.50 at 6 per cent., 
that the amount will pay a debt of $490.70 ? 

6. In what time will $740 amount to $1037.89 at 7 per cent, 
compound interest ? Ans. 5 yr. 

7. In what time will $1000 draw $171,353 at 6 per cent, com- 
pound interest? Ans. 2 yr. 8 mo 15 da. 

8. In what time will $333 amount to $376.75 at 5 p.er cent 
compound interest, payable semi-annually ? 

9. In what time will any sum double itself at 6 % compound 
interest ? At 7 % ? Ans., to last, 11 yr. 10 mo. 21 da. 

DISCOUNT. 

55G. Discount is an abatement or allowance made for the 
payment of a debt before it is due 

557. The Present Worth of a debt, payable at a future time 
without interest, is such a sum as, being put at legal interest, will 
amount to the given debt when it becomes due. 

1. What is the present worth and what the discount of $642.12 
to be paid 4 yr. 9 mo. 27 da. hence, money being worth 7 per 
cent. ? 



discount. 325 

operation. Analysis. Since $1 is the 

$1.33775, Amount of $1. present worth of $1.33775 

$642.12 -r- 1.33775 = $480 for the given time at the 

$642.12, given sum. given rate of interest, the 

480. present worth. present worth of $642.12 



$162.12, discount. must be as man y dollars as 

$1.33775 is contained times 

in $642.12. Dividing, and we obtain $480 for the present worth, and 
subtracting this sum from the given sum, we have $162.12, the dis- 
count. Hence the following 

. Rule. I. Divide the given sum or debt by the amount of $1 
for the given rate and time; the quotient will be the present worth 
of the debt 

II. Subtract the present worth from the given sum or debt ; the 
remainder will be the discount. 

Notes. — 1. The terms present worth, discount, and debt, are equivalent to 
principal, interest, and amount. Hence, when the time, rate per cent., and 
amount are given, the principal may be found by Prob. II, (553) ; and the 
mterest by subtracting the principal from the amount. 

2. When payments are to be made at different times without interest, find the 
present worth of each payment separately. Their sum will be the present worth 
< f the several payments, and this sum subtracted from the sum of the several 
payments will leave the total discount. 

EXAMPLES FOR PRACTICE. 

1. What is the present worth of a debt of $385.31, to be paid 
in 5 mo. 15 da., at 6 % ? Ans. $375. 

2. How much should be discounted for the present payment of 
a note for $429,986, due in 1 yr. 6 mo. 1 da., money being worth 
5* %? Ans. 8397.16. 

3. Bought a farm for $2964.12 ready money, and sold it again 
for $3665.20, payable in 1 yr. 6 mo. How much would be gained 
in ready mon£y, discounting at the rate of 8 % ? 

4. A man bought a flouring mill for $25000 cash ; or for $12000 
payable in 6 mo. and $15000 payable in 1 yr. 3 mo. He accepted 
the latter offer; did he gain or lose, and how much, money being 
worth to him 10 per cent. ? Ans. Gained $238.10. 

5. B bought a house and lot April 1, I860, for which he was 
to pay $1470 on the fourth day of the following September, and 

28 



§26 PERCENTAGE. 

$2816.80 Jan. 1, 1861. If he could get a discount of 10 per 
cent, for present payment, how much would he gain by borrowing 
the sum at 7 per cent., and how much must he borrow? 

6. What is the difference between the interest and the discount 
of $576, due 1 yr. 4 mo. hence, at 6 per cent. ? 

7. A merchant holds two notes against a customer, one for 
$243.16, due May 6, 1861, and the other for $178.64, due Sept. 
25, 1861 ; how much ready money would cancel both the notes 
Oct. 11, 1860, discounting at the rate of 7 % ? 

8. A speculator bought 120 bales of cotton, each bale contain- 
ing 488 pounds, at 9 cents a pound, and gave a note at 9 months 
at 8 per cent., for the amount. He immediately sold the cotton 
for $6441.60 cash, and paid his note ; how much did he gain ? 

9. Which is the more advantageous, to buy flour at $6.25 a 
barrel on 6 months, or at $6.50 a barrel on 9 months, money being 
worth 8 % ? 

10. How much may be gained by hiring money at 5 % to pay 
a debt of $6400, due 8 months hence, allowing the present worth 
of this debt to be reckoned by deducting the nominal discount ? 

Arts. $6.87+. 
BANKING. 

558. A Bank is a corporation chartered by law for the pur- 
pose of receiving and loaning money, and furnishing a paper 
circulation. 

559. A Promissory Note is a written or printed engagement 
to pay a certain sum either on demand or at a specified time. 

560. Bank Notes, or Bank Bills, are the notes made and 

issued by banks to circulate as money. They are payable in specie 

at the banks. 

Note. — A bank which issues notes to circulate as money is called a bank of 
issue; one which lends money, a bank of discount ; and one which takes charge 
of money belonging to other parties, a bank of deposit. Some banks perform 
two and some all these duties. 

561. The Maker or Drawer of a note is the person by whom 
the note is signed ; 

563* The Payee is the person to whose order the note is made 
payable; and 



BANKING- 327 

563. The Holder is the owner. 

06 4 A Negotiable Note is one which may be bought and 
sold, or negotiated. It is made payable to the bearer or to the 
order of the payee. 

5G5. Indorsing a note by a payee or holder is the act of 

writing his name on its back. 

Notes. — 1. If a note is payable to the bearer, it may be negotiated without 
indorsement. 

2. An indorsement makes the indorser liable for the payment of a note, if the 
maker fails to pay it when it is due. 

3. A note should contain the words " value received," and the sum for which 
it is given should be written out in words. 

566. The Face of a note is the sum made payable by the 
note. 

567. Days of Grace are the three days usually allowed by 
law for the payment of a note after the expiration of the time 
specified in the note. 

568. The Maturity of a note is the expiration of the days 

of grace ; a note is due at maturity. 

Note. — No grace is allowed on notes payable " on demand," without grace. 
In some States no grace is allowed on notes, and their maturity is the expira- 
tion of the time mentioned in them. 

U69. Notes may contain a promise of interest, which will 

be reckoned from the date of the note, unless some other time be 

specified. 

Note. — A note is on interest from the day it is due, even though no mention 
be made of interest in the note. 

570. A Notary, or Notary-Public, is an officer authorized 
by law to attest documents or writings of any kind to make them 
authentic. 

571. A Protest is a formal declaration in writing, made by a 
Notary-Public, at the request of the holder of a note, notifying 
the maker and the mdorsers of its non-payment. 

Notes. — 1. The failure to protest a note on the third day of grace releases the 
indorsers from all obligation to pay it. 

2. If the third day of grace or the maturity of a note occurs on Sunday or a 
holiday, it must be paid on the day previous. 

572. Bank Discount is an allowance made to a bank for the 
payment of a note before it becomes due. 



328 PERCENTAGE. 

573. The Proceeds of a note is the sum received for it when 
discounted, and is equal to the face of the note less the discount. 

57 4L* The transaction of borrowing money at banks is con- 
ducted in accordance with the following custom : The borrower 
presents a note, either made or indorsed by himself, payable at a 
specified time, and receives for it a sum equal to the face j less 
the interest for the time the note has to run. The amount thus 
withheld by the bank is in consideration of advancing money on 
the note prior to its maturity. 

Notes. — 1. A note for discount at bank must be made payable to the order 
of some person, by whom it must be indorsed. 

2. The business of buying or discounting notes is chiefly carried on by banks 
and brokers. 

575. The law of custom at banks makes the bank discount 
of a note equal to the simple interest at the legal rate, for the 
time specified in the note. As the bank always takes the interest 
at the time of discounting a note, bank discount is equal to simple 
interest paid in advance. Thus, the true discount of a note for 
$153, which matures in 4 months at 6 %, is $153— ^-go == 
$3.00, and the bank discount is $153 X .02 = $3.06. Since the 
interest of $3, the true discount, for 4 months is $3 x .02 = $.06, 
we observe that the bank discount of any sum for a given time is 
greater than true discount by the interest on the true discount 
for the same time 

Note. — Many banks take only true discount. 

CASE I. 

576. Given, the face of a note, to find the discount 
and the proceeds. 

Rule. I. Compute the interest on the face of the note for three 
days more than the specified time; the result will be the discount. 

II. Subtract the discount from the face of the note; the re- 
mainder will be the proceeds. 

Notes. — 1. When a note is on interest, payable at a future specified time, the 
amount is the face of the note, or the sum made payable, and must be made the 
basis of discount. 

2. To indicate the maturity of a note or draft, a vertical line ( | ) is used, with 
the day at which the note is nominally due on the left, and the date of maturity 
on the right; thus, Jan. 7 | , . 



BANKING. 329 

EXAMPLES FOR PRACTICE. 

1. What is the bank discount, and what are the proceeds of a 
note for $1487 due in 30 days at 6 per cent. ? 

Ans. $Discount, 8.17; Proceeds, $1478.83. 

2. What are the proceeds of a note for $384.50 at 90 days, if 
discounted at the New York Bank ? 

3. Wishing to borrow $1000 of a Southern bank that is dis- 
counting paper at 8 per cent., I give my note for $975, payable 
in 60 days; how much more will make up the required amount? 

4. A man sold his farm containing 195 A. 2 R. 25 P. for $27.50 
an acre, and took a note payable in 4 mo. 15 da. at 7 % interest. 
Wishing the money for immediate use, he got the note discounted 
at a bank; how much did he receive? Ans. $5236.169. 

5. Find the day of maturity, the term of discount, and the pro* 
ceeds of the following notes : 

$1962^ . Detroit, July 26, 1860. 

6. Four months after date I promise to pay to the order of James 
Gillis one thousand nine hundred sixty-two and y 4 ^ dollars at the 
Exchange Bank, for value received. John Demarest. 

Discounted Aug. 26, at 7%. 

Ans. Due Nov. 26 | 29 ; term of discount, 95 da.; proceeds, 
$1931.80. 

$1066/^ . Baltimore, April 19, 1859. 

7. Ninety days after date we promise to pay to the order of 
King & Dodge one thousand sixty-six and /^dollars at the Citk 
zens' Bank, for value received. Case & Sons. 

Discounted May 8, at 6 %. 

Ans. Due July 17 | 20 ; term of discount, 73 da.; proceeds, 
$1053.77. 



$784-^. Mobile, June 20, 1861. 

8. Two months after date for value received I promise to pay 
George Thatcher or order seven hundred eighty-four and T 7 ^ dol T 
lars at the Traders' Bank. Wm. Hamilton. 

Discounted July 5, at 8 %. 
28* 



330 PERCENTAGE. 



$1845/^. Chicago, Jan. 31, 1862. 

9. One month after date we jointly and severally agree to pay 
to W. H. Willis, or order, one thousand eight hundred forty-five 
and _5_o_ dollars at the Marine Bank. 

Payson & Williams. 
Discounted Jan. 31, at 2 % a month. 

Arts. Due Feb. 28 | March 3; term of discount, 31 da.; pro- 
ceeds, $1807.36. 

10. What is the difference between the true and the bank dis- 
count of $950, for 3 months at 7 per cent. ? Ans. $.29 -f. 

11. What is the difference between the true and the bank dis- 
count of $1375.50, for 60 days at 6 per cent. ? 

CASE II. 

577. Given, the proceeds of a note, to find the face. 
1. For what sum must I draw my note at 4 months, interest 
6 %, that the proceeds when discounted in bank shall be $750 ? 

operation. Analysis. We 

$1.0000 first obtain the pro- 

.0205, disc't on $1 for 4 mo. 3 da. ceeds of $1 by the 

$~9795, proceeds of $1. lastcase; then, since 

$750 -- .9795 = $765,696, Ans. $- 9795 » th f P r0 " 

ceeds of $1, $750 is 

the proceeds of as many dollars as $.9795 is contained times in $750. 

Dividing, we obtain the required result. Hence the 

Rule. Divide the proceeds by the proceeds of $iybr the time 
and rate mentioned ; the quotient will be the face of the note. 

EXAMPLES FOR PRACTICE. 

1. What is the face of a note at 60 days, the proceeds of which, 
when discounted at bank at 6 %, are $1275? Ans. $1288.53. 

2. If a merchant wishes to draw $5000 at bank, for what sum 
must he give his note at 90 days, discounting at 6 per cent. ? 

Ans. $5078.72. 

3. The avails of a note having 3 months to run, discounted at 
a bank at 7 %, were $276.84; what was the face of the note ? 



BANKING. 331 

4. James T. Fisher buys a bill of merchandise in New York at 
cash price, to the amount of $1486.90, and gives in payment his 
note at 4 months at 7^ °Jo ) what must be the face of the note ? 

5. Find the face of a 6 mo. note, the proceeds of which, dis- 
counted at 2 cf a month, are $496. Ans. $564.92. 

6. For what sum must a note be drawn at 30 days, to net 
$1200 when discounted at 5 % ? 

7. Owing a man $575, I give him a 60 day note ; what should 
be the face of the note, to pay him the exact debt, if discounted 
at I* cf a month? Ans. $593.70. 

8. What must be the face of a note which, when discounted at 
a broker's for 110 days at 1 % a month, shall give as its proceeds 
$187.50? 

CASE III. 

578. Given, the rate of bank discount, to find the 
corresponding rate of interest. 

I. A broker discounts 30 day notes at 1^ % a month; what 
rate of interest does his money earn him ? 

operation. Analysis. If we assume 

30 day notes = 33 days' time. $ 100 as the face of the 

$100 base. note, the discount for 33 

1.65, discount for 33 days. days at 1J per cent, will 

$98.35, proceeds. be $L65 and the P roceeds 

$1.65 -- .090154* =18/9 .4 oj Ans. $ 98 - 35 - We then h ave 
17 $98.35 principal, $1.65 in- 

terest, and 33 days time, to find the rate per cent, per annum, which 
we do by (549). Hence the 

Rule. I. Find the discount and the proceeds of $1 or $100 
for the time the note has to run. 

II. Divide the discount hy the interest of the proceeds at 1 per 
cent, for the same time. 

EXAMPLES FOR PRACTICE. 

1. What rate of interest is paid, when a note payable in 30 
days is discounted at 6 per cent. ? Ans. §0$- 



332 PERCENTAGE. 

2. A note payable in 2 months is discounted at 2 % a month; 
what rate of interest is paid? Arts. 5^ of . 

3. When a note payable in 90 days is discounted at \\ % a 
month, what rate of interest is paid? Ans. 6§^ff| %. 

4. What rate of interest corresponds to 5, 6, 7, 10, 12 <f dis- 
count on a note running 10 months without grace ? 

5. What rate of interest does a man pay who has a 60 day 
note discounted at |, 1, 2, 21, 3 % a month? 

CASE IV. 

579. Given, the rate of interest, to find the corres- 
ponding rate of bank discount. 

1 A broker buys 60 day notes at such a discount that his 

money earns him 2 % a month ; what is his rate % of discount ? 

operation. Analysis. If we assume 

60 da. + 3 da. = 63 da. $ 100 as the proceeds of a 

$100 base. note, the interest for 63 days 

4.20, interest for 63 da. at 24 per cent, will be $4.20, 

$10420' amount " " and tlie amount or face of 

$4.20 + 18235 = 23 i\ %, Ans. the note will be $104.20. We 
^ 1 ' then have $104.20 the prin- 

cipal, $4.20 the interest, and 63 days the time, to find the rate per 
cent., which we do by (549) as in the last case. Hence the 

Rule. I. Find the interest and the amount o/*$l or $100 for 
the time the note has to run. 

II. Divide the interest by the interest on the amount at 1 per 
cent, for the same time. 

EXAMPLES FOR PRACTICE. 

1. What rates of bank discount on 30 day notes correspond to 
5, 6, 7, 10 per cent, interest ? 

2. At what rate should a 3 months' note be discounted to pro- 
duce 8 % interest? Ans. 7§f %. 

3. At what rates should 60 day notes be discounted to pay to a 
broker 1, 1J, 2, %\ % a month? 

4. At what rate must a note payable 18 months hence, without 
grace, be discounted to produce 7 % interest? Ans. 6||| %. 



EXCHANGE. 



333 



EXCHANGE. 

580. Exchange is a method of remitting money from one 
place to another, or of making payments by written orders. 

581. A Bill of Exchange is a written request or order upon 
one person to pay a certain sum to another person, or to his order, 
at a specified time. 

582. A Sight Draft or Bill is one requiring payment to be 
made " at sight," which means, at the time of its presentation to 
the person ordered to pay^ In other bills, the time specified is 
usually a certain number of days " after sight." 

There are always three parties to a transaction in exchange, and 
usually four: 

583. The Drawer or Maker is the person who signs the 
order or bill ; 

584. The Drawee is the person to whom the order is ad- 
dressed ; 

585. The Payee is the person to whom the money is ordered 
to be paid ; and 

58©. The Buyer or Remitter is the person who purchases 
the bill. He may be himself the payee, or the bill may be drawn 
in favor of any other person.' 

587. The Indorsement of a bill is the writing upon its back, 
by which the payee relinquishes his title, and transfers the pay- 
ment to another. The payee may indorse in blank by writing his 
name only, which makes the bill payable to the bearer, and con- 
sequently transferable like a bank note ; or he may accompany his 
signature by a special order to pay to another person, who in his 
turn may transfer the title in like manner. Indorsers become sep- 
arately responsible for the amount of the bill, in case the drawee 
fails to make payment. A bill made payable to the bearer is 
transferable without indorsement. 

588. The Acceptance of a bill is the promise which the 
drawee makes when the bill is presented to him to pay it at ma- 
turity; this obligation is usually acknowledged by writing the 
word " Accepted," with his signature, across the face of the bill, 



334 PERCENTAGE. 

Notes. — 1. In this country, and in Great Britain, three days of grace are al- 
lowed for the payment of a bill of exchange, after the time specified has expired. 
In regard to grace on sight bills, however, custom is variable ; in New York. 
Pennsylvania, "Virginia, and some other States, no grace is allowed on sight bills, 

2. When a bill is protested for non-acceptance, the drawer is obligated to pay 
it immediately, even though the specified time has not expired. 

Exchange is of two kinds — Domestic and Foreign. 

589. Domestic or Inland Exchange relates to remittances 
made between different places of the same country. 

Note. — An Inland Bill of Exchange is commonly called a Draft. 

590. Foreign Exchange relates to remittances made between 
different countries. 

591# A Set of Exchange consists of three copies of the same 
bill, made in foreign exchanges, and sent by different conveyances 
to provide against miscarriage ; when one has been paid, the others 
are void. 

592. The Face of a bill of exchange is the sum ordered to 
be paid ; it is usually expressed in the currency of the place on 
which the draft is made. 

593. The Par of Exchange is the estimated value of the 
coins of one country as compared with those of another, and is 
either intrinsic or commercial. 

594:. The Intrinsic Par of Exchange is the comparative 
value of the coins of different countries, as determined by their 
weight and purity. 

595. The Commercial Par of Exchange is the comparative 
value of the coins of different countries, as determined by their 
nominal or market price. 

Note. — The intrinsic par is always the same while the coins remain un- 
changed; but the commercial par, being determined by commercial usage, is 
fluctuating. 

590. The Course of Exchange is the current price paid in 
one place for bills of exchange on another place. This price 
varies, according to the relative conditions of trade and commercial 
credit at the two places between which exchange is made. Thus, 
if Boston is largely indebted to Paris, bills of exchange on Paris 
will bear a high price in Boston. 

When the course of exchange between two places is unfavor- 



EXCHANGE. 335 

able to drawing or remitting, the disadvantage is sometimes 
avoided, by means of a circuitous exchange on intermediate places 
between which the course is favorable. 

DIRECT EXCHANGE. 

597. Direct Exchange is confined to the two places between 
which the money is to be remitted. 

508. There are always two methods of transmitting money 
between two places. Thus, if A is to receive money from B, 

1st. A may draw on B, and sell the draft; 

2d. B may remit a draft, made in favor of A. 

Note. — One person is said to draw on another person, when he is the maker 
of a draft addressed to that person. 

CASE I. 

599. To compute domestic exchange. 

The course of exchange for inland bills, or drafts, is always ex- 
pressed by the rate of premium or discount. Drafts on time, 
however, are subject to bank discount, like notes of hand, for the 
term of credit given. Hence, their cost is affected by both the 
course of exchange and the discount for time. 

1. What will be the cost of the following draft, exchange on 
Boston being in Pittsburgh at 2 J % premium? 

$600. Pittsburgh, June 12, 1860. 

Sixty days after sight, pay to William Barnard, or order, six 
hundred dollars, value received, and charge the same to our 
account. 

To the Suffolk Bank, Boston. Thomas Bauer & Co. 

OPERATION. 

$1 4- $.0225 = $1.0225, course of exchange. 

.0105, bank discount of $1, (63 da.) 

$1,012, cost of exchange for $1. 
$600 x 1.012 = $607.20, Ans. 
Analysis. From $1.0225, the course of exchange, we subtract 
$.0102, the bank discount of $1 for the specified time, and obtain 
$1,015, the cost of exchange for $1 ; then $600 X 1.012 == $607.20, the 
cost of exchange for $600. 



336 PERCENTAGE. 

2. A commission merchant in Detroit wishes to remit to his 
employer in St. Louis, $512.36 by draft at 60 days ; what is the 
face of the draft which he can purchase with this sum, exchange 
being at 1J % discount? 

OPERATION. 

$1 — $.015 = $.985, course of exchange. 
.01225, discount of $1. 

$.97275, cost of exchange for $1. 
$512.36 -4- .97275 = $532.18+, Arts. 

Analysis. From $.985, the course of exchange, we substract 
$.01225, the bank discount of $1 for the specified time, at the legal 
rate in Detroit, and obtain $.97275, the cost of exchange for $1 ; and 
the face of the draft that will cost $512.36, will be as many dollars as 
$.97275 is contained times in 512.36, which is 532.18+, timesD 
Hence we have the following 

Bule. I. To find the cost of a draft, the face being given 

Multiply the face of the draft by the cost of exchange for $1. 

II. To find the face of a draft, the cost oeing given. — Divide 

the given cost hy the cost of exchange for $1. 

Note. — The cost of exchange for $1 may always be found, by subtracting 
from the course of exchange the bank discount (at the legal rate where the draft 
is made), for the specified time. Foi sight drafts, the course of exchange is the 
gc^st of $1. 

EXAMPLES FOR PRACTICE. 

1. What must be paid in New York for a draft on Boston, at 
30 days, for $5400, exchange being at J % premium ? 

Ans. $5392.34+. 

2. What is the cost of sight exchange on New Orleans, for 
$3000, at 3 J % discount? 

3. What must be paid in Philadelphia for a draft on St. Paul 
drawn at 90 days, for $4800, the course of exchange being 
1011 % ? Ans. $4659.60. 

4. A draft was purchased for $550.62, exchange being at a 
premium of 3£ % ; what was the face ? 

5. An agent in Syracuse, N. Y., having $1324.74 due his em- 
ployer, is instructed to. remit the same by a draft drawn at 30 
days; what will be the face of the draft, exchange being at If vf 
premium? Ans. $1357.20. 



EXCHANGE. 337 

G. My agent in Charleston, S. C, sells a house and lot for 
$7500, on commission of 1} <f €j and remits to me the proceeds in 
a draft purchased at i % premium ; what sum do I receive from 
the sale of my property ? 

7. A man in Hartford, Conn., has $4800 due him in Baltimore ; 
how much more will he realize by making a draft for this sum on 
Baltimore and selling it at J % discount, than by having a draft 
on Hartford remitted to him, purchased in Baltimore for this sum 
at f % premium? Arts. $11.73. 

8. The Merchants' Bank of New York having declared a divid- 
end of 6i %, a stockholder in Cincinnati drew on the bank for the 
sum due him, and sold the draft at a premium of If %, thus real- 
izing $508.75 from his dividend; how many shares did he own? 

9. Sight exchange on New Orleans for $5000 cost $5075; 
what was the course of exchange? Arts, 1J % premium. 

10. A man in Buffalo purchased a draft on St. Paul, Minn., 
for $5320, drawn at 60 days, paying $5141.78; what was the 
course of exchange ? Ans. 2 J % discount 

CASE 11. 

600. To compute foreign exchange. 

601. The following standards of the decimal currency of the 

United States were established April 2, 1792. 

Coins. Weight. Fineness. 

Gold eagle, 270 grains, 916f thousandths. 

Silver dollar, 416 " 

Copper cent, 264 " 

In 1834, the eagle was reduced in weight to 258 grains, and in 1837 
its fineness was fixed at 900 thousandths pure, which is likewise the 
present standard of purity for all the U. S. gold and silver coins. In 
1837, also, the silver dollar was reduced in weight to 412.5 grains. 
In 1853, the silver half dollar was reduced in weight to 192 grains, 
and the smaller silver coins proportionally. 

Note. — The object of the change in the silver coinage of the United States, 
made in 1853, was to prevent its exportation by raising the nominal value of 
silver above its foreign market value. 

602. The intrinsic par of exchange between the United 
States and different countries, is given in the following 

29 



338 PERCENTAGE. 

TABLE OF FOREIGN COINS AND MONEY — Continued. 



Florin, Mecklenburg 

" Prussia and Poland 

" Tuscany , 

Franc, Belgium 

" France 

Frederick d'or, Denmark 

Gilder, Baden 

" Darmstadt, 

" Demerara 

" Frankfort 

" Netherlands 

" Wurtemberg 

Ghersh, Tripoli 

Guinea, England 

Lira, Lombardy 

" Leghorn 

" Milan 

" Venice 

Livre, Genoa 

" Leghorn 

" Switzerland 

Mark banco, Hamburg 

" current, " 

Milree, Azores 

" Brazil , 

" Madeira 

" Portugal 

Mohur, Hindostan , 

Ounce, Naples 

Pagoda, Madras 

Piaster, Tunis 

" Turkey , 

Pistareen, Spain 

Pistole, Spain 

Pound Sterling, Great Britain., 

" British Provinces 

Real, plate, Spain 

" vellon, " 

" Egypt 

Rix dollar, Austria 

" " Batavia 

" " Denmark 



Silver. 



Gold. 
Silver. 



Gold. 
Silver. 



Gold. 



Silver. 



Gold. 



Silver. 



Lower 
Denominations. 



30 groshen. 
12 soldi. 
100 centimes. 

100 « 

60 kreutzers. 
60 " 
20 stivers. 
60 kreutzers. 
20 stivers. 
60 kreutzers. 
100 paras. 
28 shillings. 
20 soldi. 
20 " 
20 " 

100 centimes. 
20 soldi. 
20 " 

100 centessini. 
16 skillings. 
16 " 
1000 reis. 
1000 « 
1000 " 
1000 " 
16 rupees. 
3 ducats. 
42 fanams. 
16 carobas. 
100 aspers. 
4 reals vellon. 

20 shillings. 
20 " 

34 marvedis. 
34 

20 piasters. 
120 kreutzers. 
48 stivers. 
96 skillings. 



3.932 



5.059 



7.109 
2.485 
1.840 



3.904 
4.861 
4.016 



.541 
.227 

.262 
.186 
.186 

.397 
.397 
.263 
.397 
.406 
.395 
.105 

.162 
.162 
.162 
.162 
.186 
.162 
.273 
.350 
.285 
.830 
.830 
1.000 
1.120 



.124 
.026 
.197 



.097 

.048 
.968 
.971 
.782 
1.051 



EXCHANGE. 339 

TABLE OF FOREIGN COINS AND MONEY. 



COINS. 



Crown, Baden 

" Bavaria 

H England 

" France 

" Geneva 

" Portugal 

" Tuscany 

" Wurtemberg 

" Zurich 

Dollar, Argentine Republic 

" Bolivia 

" Chili 

" Columbia 

" Mexico 

" Norway 

M Peru 

« Spain 

" Sweden 

Doubloon, Bolivia 

" Columbia (Bogota)... 

" " (Popayan). 

■ Chili (since 1835) 

" " (before 1835).. .. 

* La Plata 

- Mexico (average) 

" Peru (Cuzco) 

" " (Lima) 

" Spain 

Drachma, Greece 

Ducat, Austria 

" Bavaria 

" Cologne 

" Hamburg 

" Hungary 

- " Netherlands 

" Saxony 

" Sweden 

" Wurtemberg 

Florin, Austria 

" Bavaria 

" Hanover. 

" Italy 



Metal. 



Silver. 



Gold. 
Silver. 



Gold. 



Silver. 
Gold. 



Silver. 



Lower 
Denominations. 



8 reals. 

8 " 
100 cents. 
8 reals. 
8 " 
6 marks. 
8 reals. 
10 " (old). 
6 marks. 



60 batzen. 



4 gilders 
12 marks. 

60 kreutzers. 
60 « 

60 groshen. 
12 soldi. 



•Eta u> 

— «— C 00 
G O « 1-1 



5.813 



15.580 
15.617 
15.390 
15.660 
15.570 
14.660 
15.534 
15.534 
15.551 
15.570 

2.278 
2.274 
2.250 
2.257 
2.281 
2.269 
2.264 
2.267 
2.236 






1.077 
1.072 
1.100 
1.100 



1.050 
1.070 
.960 
1.016 
1.011 
1.011 
1.022 
1.005 
1.051 
1.005 
1.003 
1.059 



65 



.485 
.395 
.547 
.181 



l-?. 



"5 o > o 



1.157 
1.151 
1.181 

1.181 
1.031 

1.128 
1.148 
1.031 
1.091 
1.086 
1.086 
1.008 
1.079 
1.129 
1.079 
1.077 
1.136 



.177 



.521 
.425 
.587 
.194 



340 



PERCENTAGE. 



TABLE OF FOREIGN COINS AND MONEY — Continued. 



COINS. 



Rigsbank dollar 

Rix dollar, Norway.. 

Rouble, Russia 

Rupee, India 

Ruspone, Tuscany... 

Sequin, Tuscany 

Scudo, Milan 

" Naples 

" Rome 

" Sicily 

Sovereign, Milan 

Thaler, Brunswick... 

" Hanover 

" Hesse-Cassel. 

" Prussia 

u Saxony 

" Bremen , 

Tale, China 

" Japan 

Tomaun, Persia 

Utchlik, Tripoli 

Yirmilik, Turkey 



Silver. 



Gold. 

a 
Silver. 



Gold. 
Silver. 



Gold. 
Silver. 
Gold. 



Lower 
Denominations. 


Intrinsic par 
of U.S. gold 
coinage of 
1837. 


Intrinsic par 
of U.S. sil- 
ver coinage 
of 1837. 


Intrinsic par 
of U.S. sil- 
ver coinage 
of 1853. 


48 skillings. 




.526 


.565 


96 " 




1.051 


1.129 


100 copecks. 




.754 


.806 


16 annas. 


6.925 
2.301 


.445 


.477 


117 soldi. 




.973 


1.045 


12 carlini. 




.950 


1.021 






1.006 


1.080 


12 tari. 


6.771 


.985 


1.058 


30 groschen. 




.692 


.743 


30 « 




.694 


.735 


30 « 




.687 


.738 


30 " 




.692 


.743 


30 « 




.694 


.735 


72 grotes. 




.788 


.846 


10 mace, 100 -> 
candarines. J 




1.480 


1.590 


10 mace, 100 \ 
candarines. J 




.750 






.800 


100 mamvodis. 


2.233 






120 paras. 




.149 


.160 


20 piasters. 


.877 







Notes. — 1. The standard value of gold as compared with silver in the United 
States, is as 15.407 to 1 in the coinage of 1792, as 15.988 to 1 in the coinage of 
1837, and as 14.922 to 1 in the coinage of 1853. 

2. The relative values of gold and silver differ in the coinage of different coun- 
tries. In England the ratio is 14.288 to 1 ; in France it is 15.5 to 1 j in Ham- 
burg it is 15 to 1. 

3. In the present gold coinage of the United States, a Troy ounce of pure gold 
is equal to $20,672, and of standard gold to $18,605. In the present silver 
coinage of the United States, a Troy ounce of pure silver is equal to $1,388, and 
of standard silver to $1.25. 

GO 3* It will be seen by the table that the par of exchange 
between the United States and Great Britain is £1 = $4,861. 
Previous to the changes in the U. S. coinage, made in 1834 and 
in 1837, the par of exchange was £1 == $4.44|, or £9 = $40, 
which is called the old par of exchange. By the new par of ex- 



EXCHANGE. 341 

change, sterling money is worth about 9| % more than by the 
old par. 

604:. The course of exchange on England is usually given 
with reference to the old par of exchange. Hence, when sterling 
money is really at par, according to present standards, it is quoted 
in the market at 9| °J premium. 

605. The course of exchange between different countries may 
be expressed either by the rate per cent, above or below par, or 
by giving the sum of money in one country which is equal to a 
certain sum in another country. In the latter case, the exchange 
requires simply a reduction of currencies ; in the former, it requires 
both a reduction of currencies and a computation of percentage. 

1. What will be the cost in Boston of the following bill of ex- 
change on Liverpool, at 9| % premium? 



£432. Boston, June 16, 1860. 

At sight of this First of Exchange (Second and Third of 
same tenor and date unpaid) pay to the order of J. Simmons, 
Boston, Four Hundred Thirty-two Pounds, value received, and 
charge the same to account of 

James Lowell & Co. 
To Bichard Evans & Son, 1 
Liverpool, England. j 

Analysis. Since 

operation. exchange on Liver- 

£9 = $40 x 1 .095, course of exchange, pool is at 9J % pre- 

. 1 ' $40 x 1.005 mium, £9 will cost 

* X - 9 ' C0St 0t **» $40 x 1.095, (603) ; 

AOO $40 X 1.095 ^ 01A0 . A : and £1 will therefore 

432 x 9 =$2102.40, An*. $40 x im 

cost o . 

Multiplying the face of the bill, £432, by the cost of exchange of £1, 
we obtain $2102.40, the required cost of the bill. 

2. What is the face of a bill on London, that may be purchased 
in New York for $2768.70, exchange being at 10 % premium in* 
favor of London ? 
29* 



342 PERCENTAGE. 

OPERATION. 

£9 = $40 X 1.10, course of exchange, 

£1 = — ' — 9 cost of £1, 

40 y 1 10 

$2768.70 ^ * ' =£556 8s. 6d., Ans. 

40 X 1.10 
Analysis. We divide $2768.70, the given cost, by q * ' > ^e 

cost of exchange for £1, and obtain £556 8s. 6d., the face. 

3. What cost, in Hamburg, a bill on New Orleans for $4500, the 
course of exchange being 1 mark = $.365 ? 

operation. Analysis. Since 

$1 == i™ f a mar k ? cost f a un i t . exchange for $1 will 

$4500 X |g§ = 1232 marks 14 skillings. cost in Hamburg 

^° s of a mark, a bill 

for $4500 will cost 4500 X ££§ = 1232 marks 14 skillings. 
GOO* From these illustrations we derive the following 
Rule. I. To find the cost of a bill, the face being given. — 

Multiply the face by the cost of a unit of the currency in which the 

bill is expressed. 

II. To find the face of a bill, the cost being given. — Divide the 

given cost by the cost of a umt of the currency in which the bill is 

to be expressed. 

EXAMPLES FOR PRACTICE. 

1. What is the cost in Portland of a bill on Manchester, Eng., 
for £325 3s. 9d., at 9f % premium ? Ans. $1585.94. 

2. What must be paid in Charleston for a bill of exchange on 
Paris for 6000 francs, at 18| cents per franc? 

3. What cost in Boston a bill on St. Petersburg for 3000 roubles 
at 1J vfo premium, the par of exchange being $.754 for 1 rouble? 

4. What will be the cost in Naples of a bill of exchange on 
New York fcfr $831.12, at the rate of $.96 for 1 scudo? 

Ans. 865 scudi 9 carlini. 

5. A draft on Philadelphia cost £125 in Birmingham, Eng., 
exchange being at 8 % premium for sterling ; required the face 
of the draft* 



EXCHANGE. 343 

6. An agent in Boston, having $7536.30 due his employer in 
England, is directed to remit by a bill on Liverpool ; what is the 
face of the bill which he can purchase for this money, exchange 
being at 11 °f premium ? Arts. £1527 15s. 6d. 

7. A merchant in Cincinnati has 9087 gilders 10 stivers due 
him in Amsterdam, and requests the remittance by draft; what 
sum will he receive, exchange on U. S. being in Amsterdam at 2 J 
gilders for $1 ? 

8. A trader in London wishes to invest £2500 in merchandise 
in Lisbon ; if he remits to his correspondent at Lisbon a bill pur- 
chased for this sum, at the rate of 64. 5d. sterling per milree, what 
sum in the currency of Portugal will the agent receive ? 

Ans. 9302 milrees 325 reis. 

9. A draft on Dublin for £360 cost $1736; what was the 
course of exchange? Ans. 8 J % premium. 

10. A merchant in Baltimore, having received an importation 
of Madeira wine invoiced at 1500 milrees, allows his correspondent 
in Madeira to draw on him for the sum necessary to cover the 
cost, exchange on the United States being in Madeira 930 reis = 
$1 ; how much would the merchant have saved, by remitting a 
draft on Madeira, purchased at $1,065 per 1 milree? 

Ans. $15.40. 

11. An importer received a quantity of leghorn hats, invoiced 
at 25256 lire 16 scudi, which was paid in U. S. gold coin, ex- 
ported at a cost of 3 % for transportation and insurance, the price 
of fine gold in Leghorn being 131 lire per ounce Troy. How 
much less would the goods have cost in store, had payment been 
made by draft on Leghorn, purchased at the rate of 16 cents per 
lira? Ans. $56.78. 

_ Note. — In U. S. gold coinage, $10 contains 258 X .9 = 232.2 grains of fine 
poto, (601). 

12. When silver is worth in England 67d. per oz. fine, what 
sum of money in the U. S. silver coinage of 1853 is equal to 20 
shillings, or £1 sterling? Ans. $4,975. , 

13. At what rate of premium is Prussian coin, when $88.23 in 
U S. silver coinage of 1837 is paid for 125 thalers ? Ans. 2 %. 



344 PERCENTAGE. 

ARBITRATED EXCHANGE. 

007. Arbitration of Exchange is the process of computing 
exchange between two places by means of one or more interme- 
diate exchanges. 

Notes. — 1. When there is>only one intermediate exchange, the process is 
called Simple Arbitration ; when there are two or more intermediate exchanges, 
the process is called Compound Arbitration. 

2. The arbitrated price is generally either greater or less than the price of 
direct exchanges; and the object of arbitration is to ascertain the best route for 
making drafts or remittances. 

008. There are always three methods of receiving money 
from a place, or of transmitting money to a place, by means of 
indirect exchange through one intervening place. Thus, 

If A is to receive money from C through B, 
1st. A may draw on B, and B draw on C ; 
2d. A may draw on B, and C remit to B ; 
3d. B may draw on C, and remit to A. 

If A is to transmit money to C through B, 
1st. A may remit to B, and B remit to C ; 
2d. A may remit to B, and draw on B ; 
3d. B may draw on A, and remit to C. 

1. A man in Albany, N. Y., paid a demand in Paris of 5400 
francs, by remitting to Amsterdam at the rate of 21 cents for 10 
stivers, and thence to Paris at the rate of 28 stivers for 3 francs ) 
how much Federal money was required ? 

operation. Analysis. We are to deter- 

$(?") = 5400 francs. mine how much Federal money 

3 francs =28 stivers. is equal to 5400 francs, and the 

10 stivers = $21. question may be represented 

thus :$(?) = 5400 francs. Now 
since 3 francs = 28 stivers, and 
10 stivers = $.21, we know that 
if the required sum be multi- 
plied successively by 3 francs 
and 10 stivers, the result will be 
() = $1058.40, Ans. equal to the product of 5400 

francs by 28 stivers and $.21 
successively, (Ax. 3). Canceling the units of currency, 1 franc, 1 
stiver, and $1, and also the equal numerical factors, we have (?) 
= $1058.40, the sum required. 



(?) = $1058.40, Ans 


Or, 


( '2 


5400 
28 


10 


.21 



EXCHANGE. 345 

Or, since the course of exchange between Amsterdam and Paris 
gives 1 franc == 2 ^ stivers, and the course between Albany and Am- 
sterdam gives 1 stiver = f £ cents, we multiply the 5400 francs by 2 g 
and f J successively, using the vertical line and cancelation, and obtain 
$1058.40, as before. 

Note. — In the first statement the rates of exchange are so arranged that the 
same unit of currency shall stand on opposite sides in each two consecutive 
equations, in order that these factors may all be canceled. 

2. A resident of Naples having a bequest of $8720 made him 
in Boston, orders the remittance to be made to his agent in Lon- 
don, who remits the proceeds to Naples, reserving his commission 
of ^ % on the draft sent. If exchange on London is 9 % in 
Boston, and the rate between London and Naples is £1 for 5 
scudi, how much does the man realize from his bequest ? 

operation Analysis. "We make 

(?) scudi = $8720 the statement as in the 

$40 X 1.09 =.£9 first example, according 

£1 = 5 scudi. to the given rates of ex- 

1.005 change. Then, since the 

""(?) = 8955 scudi 3 carlini. agent k to deduct J % 

commission on the face 
of the draft before the purchase, we place 1.005 on the left as a 
divisor, (159), and obtain by cancellation 8955 scudi 3 carlini as the 
proceeds of the exchange. 

Note. — Since the par of exchange on England is £9 = $40, the course of ex- 
change will always be £9 = $40 X 1 plus the rate of exchange. 

3. A merchant in Chicago directs his agent in Albany to draw 
upon Baltimore at 1 % discount, for $1200 due from the sales 
of produce ; he then draws upon the Albany agent at 2 % pre- 
mium, for the proceeds, after allowing the agent to reserve \ % 
for his commission. What sum* does the merchant realize from 
his produce ? 

operation. Analysis. According to the 

(?)C #= _ 1200 B. given rates of exchange, 100 dol- 

100 B. = 99 A. lars in Baltimore is equal to 99 

100 A. == 102 C. dollars in Albany ; and 100 dol- 

.995 lars in Albany is equal to 102 

( ? ") = $1205.70 Arts. dollars in Chicago ; and since the 

unit of currency is the same in 
each place, being $1, we represent its exchange value in each town 



346 PERCENTAGE. 



by the initial letter, and make the statement as in the other exam- 
ples. Then, since the agent is to reserve \ fo commission from the 
avails of his draft, we place 1 — .005 = .995 on the right as a mul- 
tiplier, and obtain by cancellation ( ? ) = $1205.70, the answer. 
From these principles and illustrations we have the following 
Rule. I. Represent the required sum by ( ? ), with the proper 
unit of currency affixed, and place it equal to the given sum on the 
right. 

II. Arrange the given rates of exchange so that in any two con- 
secutive equations the same unit of currency shall stand on opposite 






III. When there is commission for drawing, place 1 minus the 
rate on the left if the cost of exchange is required, and on the right 
if proceeds are required; and when there is commission for remit- 
ting, place 1 plus the rate on the right if cost is required, and on 
the left if proceeds are required. 

IV. Divide the product of the numbers on the right by the prod- 
uct of the numbers on the left, cancelling equal factors ; the result 
will be the answer. 

Notes. — 1. Commission for drawing is commission on the sale of a draft; 
commission for remitting is commission on the purchase price of a draft. 

2. The above method is sometimes called the Chain Rule, or Conjoined Pro- 
portion. 

EXAMPLES FOR PRACTICE. 

1. A gentleman in Philadelphia wishes to deposit $5000 in a 
bank at Stockholm, by remitting to Liverpool and thence to Stock- 
holm ; if exchange on Liverpool is at 10 % premium in Phila- 
delphia, and the course between Liverpool and Stockholm is 6 
roubles 48 copecks per £1, how much money will the man have 
in bank at Stockholm, allowing the agent at Liverpool \ % for 
remitting ? Ans 6243 roubles 66 copecks. 

2. When exchange at New York on Paris is 5 francs 16 cen- 
times per $1, and at Paris on Hamburg 2£ francs per marc banco, 
what will be the arbitrated price in New York of 7680 marc 
bancos of Hamburg ? Ans. $3162.79. 

3. A man in Cleveland wishes to draw on New Orleans for a 
bank stock dividend of $750, and exchange direct on New Or- 
leans is 1} % discount; how much will he save by drawing on 



EXCHANGE. 



347 



his agent in New York at 1} % premium, allowing his agent to 
draw on New Orleans at 1 % discount, brokerage at J <f ? 

4. A gentleman in Boston drew on Wurtemberg for 6000 gild- 
ers at $.415 per gilder; how much more would he have received 
if he had ordered remittance to London, and thence to New York, 
exchange at Wurtemberg on London being 11} gilders per £1, 
and at London on New York 9} %, in favor of sterling, broker- 
age at 1} % in London for remitting? Arts. $140.46. 

5. If at Philadelphia exchange on Liverpool is at 9f % pre- 
mium, and at Liverpool on Paris 26 francs 86 centimes per £1 ; 
what is the arbitrated course of exchange between Philadelphia 
and Paris, through Liverpool? Ans. 1 franc = $.163 + - 

6. An American resident of Amsterdam wishing to obtain 
funds from the U. S- to the amount of $6400, directs his agent 
in London to draw on the U. S. and remit the proceeds to him in 
a draft on Amsterdam, exchange on the U. S. being at 8 % in 
favor of London, and the course between London and Amsterdam 
being 18d. per gilder. If the agent charges commission at J % 
both for drawing and remitting, how much better is this arbitra- 
tion than to draw directly on the U. S. at 40 cents per gilder ? 

7. A speculator in Pittsburgh, having purchased 58 shares of 
railroad stock in New Orleans, at 95 %, remits to his agent in 
New York a draft purchased at 2 % premium, with orders for the 
agent to remit the sum due in N. O. Now, if exchange on N. O. 
is at J % discount in N. Y., and the agent's commission for re- 
mitting is } %, how much does the stock cost in Pittsburgh? 

Ans. $5606.08. 

8. A banker in New York remits $3000 to Liverpool, by arbi- 
tration, as follows : first to Paris at 5 francs 40 centimes per $1 ; 
thence to Hamburg at 185 francs per 100 marcs ; thence to Am- 
sterdam at 35 stivers per 2 marcs; thence to Liverpool at 220 
stivers per £1 sterling. How much sterling money will he have 
in bank at Liverpool, and what will be his gain over direct ex- 
change at 10 % premium? 

. | Proceeds in Liverpool, £696 lis. 2d. 
US ' \ Gain by arbitration, £82 16s. 5d. 



348 PERCENTAGE. 

EQUATION OF PAYMENTS. 

609. Equation of Payments is the process of finding the 
mean or equitable time of payment of several sums, due at dif- 
ferent times without interest. 

610. The Term of Credit is the time to elapse before a debt 
becomes due. 

611. The Average Term of Credit is the time to elapse before 
several debts, due at different times, may all be paid at once, with- 
out loss to debtor or creditor. 

613. The Equated Time is the date at which the several 
debts may be canceled by one payment. 

613. To Average an Account is to find the mean or equit- 
able time of payment of the balance. 

614. A Focal Date is a date to which all the others are com- 
pared in averaging an account. 

Note. — Each item of a book account draws interest from the time it is due, 
which may be either at the date of the transaction, or after a specified term of 
credit 

In averaging, there are two kinds of equations, Simple and 
Compound. 

619. A Simple Equation is the process of finding the aver- 
age time when the payments or account contains only one side, 
which may be either a debit or credit. 

616. A Compound Equation is the process of averaging 
when both debts and credits are to be considered. 

SIMPLE EQUATIONS. 
CASE I. 

617. When all the terms of credit begin at the same 
date. 

1. In settling with a creditor On the first day of April, I find 
that I owe him $12 due in 5 months, $15 due in 2 months, and 
$18 due in 10 months ; at what time may I pay the whole amount? 



EQUATION OF PAYMENTS. 349 

operation. Analysis. The 

$12 x 5 = 60 "whole amount to be 

15 X 2 = 30 paid, as seen in the ope- 

18 X 10 = 180 ration, is $45 ; and we 

a 45 270 are to ^ n< ^ now long it 

270 -r- 45 = 6 mo., average credit, sha11 De withheld, or 
Apr. 1, + 6 mo. = Oct. 1, Arts. what term of credit it 

shall have, as an equiv- 
alent for the various terms of credit on the different items. Now 
the value of credit on any sum is measured by the product of the 
money and time. Therefore, the credit on $12 for 5 mo. = the credit 
on $60 for 1 mo., because 12 X 5 = 60 X 1. In like manner, we have 
the credit on $15 for 2 mo. = the credit on $30 for 1 mo. ; and the 
credit on $18 for 10 mo. = the credit on $180 for 1 mo. Hence, by 
addition, the value of the several terms of credit on their respective 
sums equals a credit of 1 month on $270 ; and "this equals a credit of 
6 months on $45, because 45 X 60 = 270 X 1. Hence the following 

Rule. I. Multiply each payment by its' term of credit, and 
divide the sum of the products by the sum of the payments; the 
quotient will be the average term of credit. 

II. Add the average term of credit to the date at which all 
the credits begin ; the result will be the equated time of payment. 

Notes. — 1. The periods of time used as multipliers must all be of the same 
denomination, and the quotient will be of the same denomination as the terms 
of credit ; if these be months, and there be a remainder after the division, con- 
tinue the division to days by reduction, always taking the nearest unit in the last 
result. 

2, The several rules in equation of payments are based upon the principle of 
bank discount; for they imply that the discount of a sum paid before it is due 
equals the interest of the same amount paid after it is due. 

EXAMPLES FOR PRACTICE. 

1. On the first day of January, 1860, a man gave 3 notes, the 
first for $500 payable in 30 days ; the second for $400 payable in 
60 days; the third for $600 payable in 90 days. What was the 
average term of credit, and what the equated time of payment ? 

Ayis, Term of credit, 62 da.; time of payment, Mar. 3, 1860. 

2. A man purchased real estate, and agreed to pay i of the price 
in 3 mo., i in 8 mo., and the remainder in 1 year. Wishing to 
cancel the whole obligation at a single payment, how long shall 
this payment be deferred ? 



850 



PERCENTAGE. 



3. I owe $480 payable in 90 days, and $320 payable in 60 days. 
My creditor consents to an extension of time for 1 year, and offers 
to take my note for the whole amount on interest at 6 per cent, 
from the equated time, or a note for the true present worth of 
both debts, on interest from date. How much will I gain if I 
choose the latter condition? Ans. $1.10 +. 

4. Bought merchandise April 1, as follows: $280 on 3 mo., 
$300 on 4 mo., $200 on 5 mo., $560 on 6 mo.: what is the 



equated time of payment? 



Ans. Aug. 15. 



CASE II. 

618. When the terms of credit begin at different 
dates. 

1. When does the amount of the following bill become due, 
per average ? 

Charles Crosby, 

1860. To Bronson & Co., Dr. 

Jan. 12. To Mdse., $400 

" 16. " Mdse. on 2 mo., 600 

Apr. 20. " Cash, 375 

FIRST OPERATION. SECOND OPERATION. 



Due 


da. 

64 
99 


Items. 

400 
600 
375 

1375 


Prod. 


Jan. 12 
Mar. 16 
Apr. 20 


38400 
37125 






75525 



Due. 


da. 

99 

35 




Items. 

400 
600 
375 

1375 


Prod. 


Jan. 12 
Mar. 16 
Apr. 20 


39600 
21000 






60600 



Ans 



75525 ~ 1375 == 55 da. 
55 da. after Jan. 12, 
or Mar. 7. 



60600 -r 1375 = 44 da. 

A { 44 da, before Apr. 20, 
1 or Mar. 7. 



Analysis. The three items of the bill are due Jan. 12, Mar. 16, 
and Apr. 20, respectively. In the first operation we use the earliest 
maturity, Jan. 12, for a focal date, and find the difference in days 
between this date and each of the others ; thus, from Jan. 12 to Mar. 






EQUATION OF PAYMENTS. 351 

16 is 64 da. ; from Jan. 12 to Apr. 20 is 99 da. Hence, from Jan. 12 
the first item has no credit, the second has 64 days' credit, and the 
third 99 days' credit, as appears in th # e column marked da. We now 
proceed to find the products as in Case I, whence we obtain the ave- 
rage credit, 55 da., and the equated time, Mar. 7. 

In the second operation, the latest maturity, Apr, 20, is taken for a 
focal date, and the work may be explained thus ; Suppose the account 
to be settled Apr. 20. At that time the first item has been due 99 
days, and must therefore draw interest for this time. But interest 
on $400 for 99 days == the interest on $39600 for 1 day. The second 
item must draw interest 35 days ; but interest on $600 foi 35 days = 
interest on $21000 for 1 day. Taking the sum of the products, we find 
that the whole amount of interest due Apr. 20 equals the interest on 
$60600 for 1 day ; and this is found, by division, equal to the interest 
on $1375 for 44 da., which is the average term of interest. Hence 
the account would be settled Apr. 20, by paying $1375, with interest 
on the same for 44 days. This shows that the $1375 has been used 
44 days, that is, it falls due Mar 7, without interest. Hence we have 
the following 

Rule. I. Find the time at which each item becomes due, by 
adding to the date of each transaction the term of credit, if any be 
specified , and write these dates in a column. 

II. Assume either the earliest or the latest date for a focal date, 
and find the difference in days between the focal date and each of 
the other dates, and write the results in a second column. 

III. Write the items of the account in a third column, and mul- 
tiply each by the corresponding number of days in the preceding 
column, writing the products in a fourth column. 

IV. Divide the sum of the products by the sum of the items. 
The quotient will be the average term of credit or interest, and 
must be reckoned from the focal date TOWARD the other dates, to 
find the equated time of payment. 

N 0TE . — When dollars and cents are given, it is generally sufficient to take 
only dollars in the multiplicand, rejecting the cents when less than 50, and car- 
rying 1 to the dollars, if the cents are more than 50. 



352 PERCENTAGE. 

examples tor practice. 

1. James Gordon, 

1860. To Henry Lancey, Dr. 
Mar. 4. To 100 yd. Cassimere, @ $2 50, $250 

« 25. " 300C " French Prints," .12, 360 

Apr. 16. " 1200 " Sheeting, " .08, 96 

" 30. " 400 " OilCloth, " .50, 200 

May 17. " Sundries, 350 

When is the above bill due, per average ? 

Ans. Apr. 12, 1860. 

2. I sell goods to A at different times, and for different terms 
of credit, as follows : 

Sept. 12, 1859, a bill on 30 days' credit, for $180 

Oct. 7, " " 30 

Nov. 16, " " 60 

Dec. 20, « " 90 

Jan. 25, 1860, " 30 

Feb. 24, " " 30 

If I take his note in settlement, at what time shall interest 
commence ? 

3. What is the average of the following account ? 

1860, Oct. 1. Mdse., on 60 da., $240 

" Nov. 12. " " " 500 

" Dec. 25. " " " 436 

1861, Jan. 16. " " " 325 

" Feb. 24. " " " 436 

" Mar.17. " " " 537 

Ans. Feb. 7, 1861. 

4. I have 4 notes, as follows: the first for $350, due Aug. 16, 
1859 - the second for $250, due Oct. 15, 1859 ; the third for $300, 
due Dec. 14, 1859; the fourth for $248, due Feb. 12, 1860. 
When shall a note for which I may exchange the four, be made 
payable ? 



a a 


300 


u u 


150 


ti « 


350 


u a 


130 


a it 


140 



EQUATION OF PAYMENTS. 



353 



Dr. 



COMPOUND EQUATIONS. 

<519. 1. Average the following account. 
John Lyman. 



Cr. 



1860. 








1860. 








June 12 


To Mdse. 


530 


00 


June 25 


By draft at 30 da. 


480 


00 


Sept. 12 


a a 


428 


00 


Aug. 20 


" cash, 


280 


00 


Oct. 28 


" Sundries, 


440 


00 


Oct. 8 


a u 


140 


00 



OPERATION. 



Dr. 



Cr. 



Due. 


Da. 


Items. 


Products. 


Due. 


Da. 


Items. 


Products. 


June 12 

Sept. 12 
Oct. 28 


138 

46 




530 

428 
440 


73140 

19688 


July 28 
Aug. 20 
Oct. 8 


96 

69 
20 


480 
230 
140 


44640 

15870 

2800 




1398 
850 


92828 
63310 






850 


63310 








Balances, 


548 


29518 





Focal { 
date, I 



29518 ~ 548 = 54 da., average term of interest. 
Oct. 28 — 54 da. = Sept. 4, balance due. 

Analysis. — In this operation we have written the dates of maturity 
on either side, allowing 3 days' grace to the draft. The latest date, 
Oct. 28, is assumed as the focal date for both sides, and the two columns 
marked da. show the difference in days between the focal date and 
each of the other dates. The products are obtained as in simple 
equations, and the balance found between the items on the two sides, 
and also between the products. These balances, being both on the 
Dr. side, show that there is due on the day of the focal date, $548, 
with interest on $29518 for 1 day. By division, this interest is found 
to be equal to the interest on $548 for 54 days. Hence this balance, 
$548, has been due 54 days ; and reckoning back from the focal date, 
we obtain the equated time of payment, Sept. 7. 

Had we taken the earliest maturity, June 12, for the focal date, we 
should have obtained 84 days for the interval of time ; and since in 
this case the products would represent the credit to which the several 
items are entitled after June 12, we should add 84 days to the focal 
date, which would give Sept. 4, as before. 

2. When is the balance of the following account due, per 
average ? 

30* X 



354 



PERCENTAGE. 
Charles Derby. 



Dr. 














Cr. 


1859. 
Jan. 21 
Mar. 5 

■ " 22 


To Mdse. 

u u 


32 
145 
194 


00 
00 

00 


1859. 
Jan. 1 
Feb. 4 
Mar. 30 


By cash, 

u u 

U It 


84 
40 
12 


00 
00 

00 



OPERATION. 



Dr. 



Cr. 



Due. 


Da. 


Items. 


Products. 


Due. 


Da. 


Items. 


Products. 


Jan. 21 
Mar. 5 

" 22 


68 

25 

8 


32 
145 
194 


2176 
3625 
1552 


Jan. 1 
Feb. 4 
Mar. 30 


88 

54 




84 
40 
12 


7392 

2160 




371 
136 


7353 




136 


9552 
7353 










Balance of account, 


235 


Balance of products, 


2199 



2199 -T- 235 = 9 da. ; Mar. 30+9 da. = Apr. 8, Ans. 
Analysis. We take the latest maturity, Mar. 30, for the focal date, 
and consequently the products represent the interest due upon the 
several items, at that date. We find the balance of the items upon 
the Dr. side, and the balance of the products upon the Cr. side. The 
debtor therefore owes, on Mar. 30, $235, but is entitled to such a term 
of interest on the same as will be equivalent to the interest on $2199 
for 1 day, which by division, is found to be 9 da. Hence the balance 
is due Mar. 30+9 da. = Apr. 8. Thus we see that when the balances 
are on opposite sides, the interval of time is counted from the other 
dates. If we take, in this example, the earliest date for the focal date, 
the balances will both be upon the Dr. side, and the interval of time 
will be 97 da., which reckoned forward from the focal date, will give 
the equated time as before. 

620. From these examples we derive the following 
Rule. I. Find the time when each item of the account is due, 
and write the dates } in tico columns, on the sides of the account to 
which+they respectively belong. 

II. Use either the earliest or the latest of these dates as the focal 
date for both sides, ana find the products as in the last case. 

III. Divide the balance of the products by the balance of the 
account ; the quotient will be the interval of time, which must be 
reckoned from the focal date TOWARD the other dates when both 



EQUATION OF PAYMENTS. 



355 



balances are on the same side of the account, but PROM the other 
dates when the balances are on opposite sides of the account. 

Notes. — 1. Instead of the products, we may obtain the interest, at any per 
cent., on the several items for the corresponding intervals of time, and divide 
the balance of interest by the interest on the balance of the account for 1 day; 
the quotient will be the interval of time to be added to, or subtracted from the 
focal date, according to the rule. The time obtained will be the same, at what- 
ever rate the interest be computed. 

2. There may be such a combination of debits and credits, that the equated 
time will be earlier or later than any date of the account. 



EXAMPLES FOR PRACTICE. 

1. Required, the average maturity of the following account. 
A. Z. Armour. 



Dr. 














Cr. 




1859. 








1859. 








Feb. 12 


To Mdse. 


85 


75 


March 15 


By bal. old acc't. 


97 


36 


" 25 


a a 


36 


24 


April 17 


" cash, 


56 


00 


April 16 


a a 


174 


96 


May 25 


n a 


25 


00 


May 20 


Ct it 


94 


78 


June 8 


" sundries, 


94 


75 



OPERATION. 



Dr. 














Cr. 


Due. 


Da. 


Items. 


Int. 


Due. 


Da. 


Items. 


Int. 


Feb. 12 

25 

April 16 

May 20 


116 

103 

53 

19 


85.75 

36.24 

174.96 

94.78 


1.66 

.62 

1.53 

.30 


March 15 
April 17 
May 25 
June 8 


88 
52 
14 


97.36 
56.00 
25.00 
94.75 


1.43 

.49 
.06 




391.73 
273.11 


4.11 
1.98 






273.11 


1.98 








Balances, 


118.62 ' 


2.13 





Int on $118.62 for 1 da. = $.0198. 
2.13-^.0198=108 da.; June 8—108 da.=Feb. 20,1859, Ans. 

Analysis. Taking the latest maturity, June 8, for the focal date, 
we find the interest of each item, at 6 fo, from its maturity to the 
focal date ; then, taking the balance, we find the interest due on the 
account to be §2.13. Dividing this interest by the interest on the 
balance of the items for 1 day, we obtain 108 da., the time required 
for the interest, $2.13, to accrue. The average maturity, therefore, 
is June 8 — 108 da. = Feb. 20, 1859. 

It is evident that when the balances occur on opposite sides, the 
interval of time will be reckoned as in the method by products. 



356 



PERCENTAGE. 



2. What is the balance of the following account, and when is 

it due ? 

Thomas Lardner. 

Dr. Cr. 



I860. 


• 






1860. 








March 1 


To Sundries, 


436 


00 


March 25 


By draft, at 60 da. 


400 


00 


April 12 


" Md.se. 


548 


00 


April 6 


u a 3Q a 


650 


00 


July 16 


a a 


312 


00 


June 20 


" cash, 


200 


00 


Sept. 14 


a u 


536 


00 


Aug. 3 


a u 


84 


00 



Ans. Balance, $498 ; due June 30, 1860. 
3. When shall a draft for the settlement of the following ac- 
count be made payable ? 

David Sanford. 



Dr. 














Cr, 


1859. 
Jan. 1 
Feb. 12 
March 16 
June 25 


To Mdse. on 3 mo. 

a a a 2 " 

" Sundries, 
" Mdse. 


54 
28 
95 
26 


36 
45 
75 
32 


1859. 
April 1 
May 16 
June 12 

" 20 


By cash, 
" draft, at 30 da. 

a a 

" cash, 


50 

30 

125 

150 


00 
00 
00 
00 


Dr. 




Ans. Aug. 28, 1858. 

Oliver Wainwright. 

Cr. 


1858. 
Jan. 1 
Feb. 1 
March 17 
April 1 


To Mdse. 

a a 
it u 
a it 


36 

48 
72 
98 


72 
25 
36 

48 


1858. 

Jan. 10 

« 21 

March 23 

April 6 


By cash, 

u a 

" sundries, 
a tt 


98 

25 

15 

8 


72 
84 
17 
96 



If the above account were settled April 6, 1858, by draft on 
time, how many days' credit should be given ? Ans. 20 da. 

4. I owe $1000 due Apr. 25. If I pay $560 Apr. 1, and $324 
Apr. 21, when, in equity, should I pay the balance ? 

Ans. July 30. 

Note. — Make the $1000 the Dr. side of an account, and the payments the Cr. 
side, and then average. 

5. A man owes $684, payable Aug. 12, and $468, payable Oct. 
15. If he pay $1000 Aug. 1, what will be the equated time for 
the payment of the balance ? Ans. Dec. 15. 

6. A man holds 3 notes, the first for $500, due March 1, the 
second for $800, due June 1, and the third for $600, due Aug. 1. 
He wishes to exchange them for two others, one of which shall 
be for $1000, payable Apr. 1 ; what shall be the face and when 
the maturity of the other ? 

Ans. Face, $900 ; maturity, June 14. 



EQUATION OF PAYMENTS. 



357 



7. A owes $500, due Apr. 12, and $1000, due Sept. 20, and 
wishes to discharge the obligation by two equal payments, made 
at an interval of 60 days ; when must the two payments be made ? 

Ans. 1st, June 28 ; 2d, Aug. 27. 

8. When shall a note be made payable, to balance the following 

account ? 

James Tyler. 



Dr. 
















Cr. 


1859. 










1859. 








June 12 


To Mdse. on 3 mo. 


530 


84 


Sept. 14 


By cash, 


436 


00 


" 20 


a u 


t a 


236 


48 


" 25 


a a 


320 


00 


" 30 


si a 


t it 


739 


56 


Oct. 3 


a a 


560 


00 


July 5 


a fe 


t a 


273 


44 


" 17 


u tt 


370 


00 


" 16 


u a 


i a 


194 


78 


Nov. 16 


a « 


840 


00 


« 29 


a u 


t U 


536 


42 


ft 24 


a a 


560 


00 



9. I received goods from a wholesale firm in New York, in 
parcels, as per bills received, namely : Apr. 1, a bill for $536.78 ; 
May 16, $2156.94; June 12, $843.75; July 12, $594.37; Sept. 
18, $856.48. In part payment, I remitted cash as follows : June 
3, $500; July 1, $1000; Nov. 1, $1500. When is the balance 
payable, allowing credit of 2 months for the merchandise ? 

Ans. June 17. 



ACCOUNT SALES. 

631* An Account Sales is an account rendered by a commis- 
sion merchant of goods sold on account of a consignor, and con- 
tains a statement of the sales, the attendant charges, and the net 
proceeds due the owner. 

633. Guaranty is a charge made in addition to commission, 
for securing the owner against the risk of non-payment, in case of 
goods sold on credit. 

633. Storage is a charge made for keeping the goods, and 
may be reckoned by the week or month, on each article or piece. 

634:. Primage is an allowance paid by a shipper or consignor 
of goods to the master and sailors of a vessel, for loading it. 

625. A commission merchant having sold a shipment of 
goods by parts at different times, and on various terms, makes a 
final settlement by deducting all charges, and accrediting the owner 
with the net proceeds. It is evident, therefore, 



358 



PERCENTAGE. 



I. That commission and guaranty should be accredited to the 
agent at the average maturity of the sales. 

II. That the net proceeds should be accredited to the con- 
signor at the average maturity of the entire account. 

Hence the following 

Rule. I. To compute the storage. — Multiply each article or 
parcel by the time it is in store, and multiply the sum of the pro- 
ducts by the rate per unit ; the result will be the storage. 

II. To find when the net proceeds are due. — Average the sales 

alone, and the result will be the date to be given to the commission 

and guaranty ; then make the sales the Cr. side, and the charges 

the Dr. side, and average the entire account by a compound equation. 

Note. — In averaging, either the product method or the interest method may- 
be used. 

EXAMPLES FOR PRACTICE. 

1. Account sales of 100 pipes of gin, received per ship Hispan- 
iola, from Havana, on a|c. of Tyler, Jones & Co. 



I860 




April 


15 


May 


5 


June 


.28 


April 


1 


a 


1 


u 


1 


June 


28 



Sold 32 Pipes, 4160 gal. @ $1.05, on 30 days,.. 

" 40 " 5240 " @ 1.02, cash, , 

" 28 " 3650 " @ 1.00, " , 



100 



CHARGES. 



To Freight and Primage, $136.76 

" Wharfage and Cartage, 48.54 

" Duty Bonds, at 60 days, 3207.07 

" Storage from April 1, viz. : 

On 32 Pipes, 2 wks 64 wks. 

" 40 " 5 " ... 200 " 
" 28 " 13 " ... 364 " 

100 " equal to 628 "@6c 37.68 

" Commission on $13362.80. at 2% % 334.07 

" Guaranty on $4368, at 2^ % 109.20 



4368 
5344 
3650 



13362 



3873 



00 
80 
00 



80 



32 



What are the net proceeds of the above account, and when due ? 
Ans. Net proceeds, $9489.48; due, May 24, 1860. 

Note. — The time for which storage is charged on each part of the shipment 
is the interval, reduced to weeks, between Apr. 1, when the pipes were received 
into store, and the date of sale. Every fraction of a week is reckoned a full week. 

2. A commission merchant in Boston received into his store on 

May 1, 1859, 1000 bbl. of flour, paying as charges on the same 



EQUATION OF PAYMENTS. 



359 



day, freight $175.48, cartage $56.25, and cooperage $8.37. He 
sold out the shipment as follows: June 3, 200 bbl. @ $6.25; 
June 30, 350 bbl. @ $6.50; July 29, 400 bbl. @ $6.12*; Aug. 
6, 50 bbl. @ $6.00. Required the net proceeds, and the date 
when they shall be accredited to the owner, allowing commission 
at 3 J ^, and storage at 2 cents per week per bbl. 

Arts. Net proceeds, $5614.28 ; due, July 10. 



SETTLEMENT OF ACCOUNTS CURRENT. 

626. To find the cash balance of an account current, 
at any given date. 

J, Burns in account current with Tyler & Co. 



Dr. 



Cr. 



1860. 








1860. 








Feb. 25 


To Mdse. on 3 mo. 


360 


75 


March 1 


By cash on acct. 


250 


00 


March 20 


" « " 3 " 


240 


56 


April 20 


" accept, at 30 da. 


300 


00 


April 26 


u a a 3 a 


875 


24 


June 12 


" Sundries, 


375 


00 


June 24 


a a « 2 " 


235 


25 


" 27 


" cash on acct. 


400 


00 



I . 



Kequired the cash value of the above account, July 1, 1860, 
interest at 6 %. 



OPERATION. 



Dr. 



Cr. 



Due. 


Da. 


Items. Int. 


Cash val 


Due. 


Da. 


Items. Int. 


Cash val. 


May 25 
June 20 
July 26 
Aug. 24 


37 
11 

25 

54 


360.75 + 2.22 
240.56 -f .44 
875.24 — 3.64 
235.25—2.12 


362.97 
241.00 
871.60 
233.13 


March 1 

May 20 

June 12 

« 27 


122 

42 

19 

4 


250.00 + 5.08 
300.00 -f- 2^10 
375.00 + L19 
400.00+ .27 


255.08 
302.10 
376.19 
400.27 








1708.70 








1333.64 



$1708.70 —$1333.64 = $375.06. Ans. 

Analysis. For either side of the account we write the dates at 
which the several items are due, and the days intervening between 
these dates and the day of settlement, July 1. We then compute the 
interest on each item for the corresponding interval of time, and add 
it to the item if the maturity is before July 1, and subtract it from 
the item if the maturity is after July 1 ; the results must be the cash 
values of the several items on July 1. Adding the two columns of 
cash values, and subtracting the less sum from the greater, we have 
$386.58, the cash balance required. Hence the 



360 



PARTNERSHIP. 



Rule. I. Find the number of days intervening between each 
maturity and the day of settlement. 

II. Compute the interest on each item for the corresponding 
interval of time ; add the interest to the item if the maturity is 
before the day of settlement, and subtract it from the item if the 
maturity is after the day of settlement ; the results will be the cash 
values of the several items. 

III. Add each column of cash values, and the difference of the 
amounts will be the cash balance required. 



EXAMPLES FOR PRACTICE. 

1. Find the cash balance of the following account for June 1, 
1858, interest at 6 per cent. ? 

Alvan Pafke. 



Dr. 














Cr. 


1858. 








1858. 








Jan. 12 


To check, 


500 


36 


Jan. 1 


By bal. from old acct. 


536 


72 


" 26 


a a 


250 


48 


Feb. 3 


" cash, 


486 


57 


Feb. 13 


it tt 


400 


00 


March 26 


a a 


1260 


78 


March 16 


a a 


750 


00 


April 20 


a ii 


756 


36 


April 25 


a a 


200 


00 


May 12 


a it 


248 


79 



Ans. $1202.11. 

2. What is the cash balance of the following account on Dec. 

31, at 7 per cent. ? 

James Hanson. 



Dr. 














Cr. 




1859. 








1859. 








Sept. 3 


To Sundries, 


478 


36 


Sept. 17 


By Sundries, 


96 


54 


Oct. 2 


" Mdse. on 3 mo. 


256 


37 


" 20 


" cash on acct. 


200 


00 


" 21 


a a a 3 a 


375 


26 


Oct, 3 


tt a a 


325 


00 


Nov. 12 


a a ti 3 a 


80 


00 


Nov. 17 


a a a 


50 


00 


Dec. 15 


" Sundries, 


148 


76 


Dec. 27 


a u u 


84 


00 



PARTNERSHIP. 

627. Partnership is a relation established between two or 
more persons in trade, by which they agree to share the profits 
and losses of business according to the amount of capital furnished 
by each, and the time it is employed. 

G28. The Partners are the individuals thus associated. 

Notr. — The terms Capital or Stocky Dividend, and Assessment, have the same 
signification in Partnership as in Stocks. 



PARTNERSHIP. 3gj 

CASE I. 

629. To find each partner's share of the profit or 
loss, when their capital is employed for equal periods 
of time. 

1. A and B engage in trade; A furnishes $500, and B $700 as 
capital ; they gain $96 ; what is each man's share ? 

operation. Analysis. The whole 

$ 5vv amount of capital em- 

$ 700 ployed is $500 + $700 

$1200, whole stock. =$1200 ; hence, A fur- 

/ 2 0_o_ ^ _5^ A > s part of the stock . nishes T %%% = ^ of the 

7 00_ = 3 B ? s " " u u capital, and B furnishes 

$96 °x T 4 = $40 A's share of the gain. A% = 12 of the capi- 

$96 x {L == $56, B's " " " « tal * And smce each 

man's share of the pro- 
fit or loss will have the same ratio to the whole profit or loss as his 
part of the capital has to the whole capital, A will have ^ of the 
$96, and B f% of the $96, for their respective shares of the profits. 

We may also regard the whole capital as the first cause, and each 
man's share of the capital as the second cause, the whole profit or loss 
as the first effect, and each man's share of the profit or loss as the 
second effect, and solve by proportion thus : 

1st cause. 2d cause. 1st effect. 2d effect. 

$1200 : ' $500 = $96 : ( ? ) = $40, A's gain, 
$1200 : $700 = $96 : (?) = $56, B's « 
Hence we have the following 

Rule. Multiply the whole profit or loss by the ratio of each 
man's share of the capital to the whole capital. Or, 

The whole capital is to each man's share of the capital as the 
whole profit or loss is to each man's share of the profit or loss 

EXAMPLES FOR PRACTICE. 

1. Three men engage in trade; A puts in $6470, B $3780, and 
C $9860, and they gain $7890. What is each partner's share of the 
profit? Ans. A's, $2538.453; B's, $1483.053; C's, $3868.493. 

2. B and C buy pork to the amount of $1847.50, of which B 

pays $739, and C the remainder. They gain $375 ; what is each 

one's share of the gain? 
31 



362 PARTNERSHIP. 

3. A, B, and C form a company for the manufacture of woolen 
cloths. A puts in $10000, B $12800, and C $3200. C is allowed 
$1500 a year for personal attention to the business; their ex- 
penses for labor, clerk hire, and other incidentals for 1 year are 
$3400, and their receipts during the same time are $9400. What 
is A's, B's, and C's income respectively from the business ? 

4. Four persons rent a farm of 115 A. 32 P. at $3.75 an acre. 
A puts on 144, B 160, C 192, and D 324 sheep; how much rent 
ought each to pay ? 

5. Three persons gain $2640, of which B is to have $6 as often 
as C $4, and as often as D $2 ; how much is each one's share ? 

6. Six persons are to share among them $6300 ; A is to have 
^ of it, B i, C |, D is to have as much as A and C together, and 
the remainder is to be divided between E and F in the ratio of 
3 to 5. How much does each receive ? 

Ans. A, $900; B, $1260; C, $1400; 
D, $2300; JE, $165; F, $275. 

7. Two persons find a watch worth $90, and agree to divide the 

value of it in the ratio of | to | ; how much is each one's share ? 

Note. — If the fractions be reduced to a common denominator, they will be to 
each other as their numerators, (418, III). 

8. A father divides his estate worth $5463.80 between his two 
sons, giving the elder J more than the younger ; how much is each 
son's share? Ans. Elder, $2892.60; younger, $2571.20. 

9. Three men trade in company. A furnishes $8000, and B 
$12000 Their gam is $1680, of which C's share is $840; required, 
C's stock, and A's and B's gain. Ans. C's stock, $20,000. 

10. Four persons engage in the lumber trade, and invest jointly 
$22000; at the expiration of a certain time, A's share of the 
gain is $2000, B's $2800.75, C's $1685 25, and D's $1014; how 
much capital did each put in ? Ans. I) put in $3042. 

11. A legacy of $30,000 was left to four heirs in the propor- 
tion of ^, |, |, and ^, respectively; how much was the share of 
each? 

12. Three men purchase a piece of land for $1200, of which 
sum C pays $500. They sell it so as to gain a certain sum, of 



PARTNERSHIP. 3g3 

which A takes $71.27, and B $142.54; how much do A and B 
pay, and what is C's share of the gain t Ans. C's gain, $152.70. 

13. Three persons enter into partnership for the manufacture 
of coal oil, with a joint capital of $18840. A puts in $3 as often 
as B puts in $5, and as often as C puts in $7. Their annual gain 
is equal to C's stock; how much is each partner's gain? 

14. A, B, and C are employed to do a piece of work for $26.45. 
A and B together are supposed to do f of the work, A and C T 9 ^, 
and B and C ^|, and are paid proportionally ; how much must 
each receive? Ans. A, $11.50; B, $5.75; C, $9.20. 

CASE II. 

630. To find each partner's share of the profit or loss 
when their capital is employed for unequal periods of 
time. 

It is evident that the respective shares of profit and loss will 
depend equally upon two conditions, viz. : the amount of capital 
invested by each; and the time it is employed. Hence they will 
be proportional to the products of these two elements. 

1. Two men form a partnership ; A puts in $320 for 5 months, 
and B $400 for 6 months. They lose $140 ; what is each man's 
share of the loss ? 

OPERATION. 4 

$320 X 5 = $1600, A's capital for 1 mo. 
$400 + 6 = $2400, B's " " " 

$4000, entire " " " 
$16M = | ? A's share in the partnership. 

$2400 3 "R'q it ti tt 

4 000 — 5> - b 

$140 x| = $56, A's loss. 
$140 x | = $84, B's loss. 

Analysis . The use of $320 for 5 months is the same as the use of 
5 times $320, or $1600, for 1 month ; and the use of $400 for 6 months 
is the same as the use of 6 times $400, or $2400, for 1 month ; hence 
the use ot the entire capital is the same as the use of $1600 + $2400 
s=a $4000 for 1 month. A's interest in the partnership is therefore 
!£§S = !> and he will suffer § of the loss, or $140 X § = $56 ; and 



364 PARTNERSHIP. 

B's interest in the partnership is f ££J — § , and he will suffer § of the 

loss, or $140 X § = $84. 

We may also solve by proportion, the causes being compounded of 

the two elements, capital and time • thus : 

$4000 : $1600 = SI 40 : (?) = $56, A's loss 
$4000 : $2400 = $140 : (?) = $84, B's loss. 

Hence the following 

Rule, Multiply each ma?i's capital by the time it is employed 
in trade, and add the products. Then multiply the entire profit or 
loss hy the ratio of eacji product to the sum of the products ; the 
results will be the respective shares of pro jit or loss of each part- 
ner. Or, 

Multiply each maris capital by the time it is employed in trade, 
and regard each product as his capital, and the sum of the pro- 
ducts as the entire capital, and solve by proportion, as in Case I. 

EXAMPLES FOR PRACTICE. 

1. A, B, and C enter into partnership. A puts in $357 for 5 
months, B $371 for 7 months, and C $154 for 11 months, and they 
gain $347.20; how much is each ones share? 

Ans. A's $102; B's $148.40; C's $96.80. 

2. Three men hire a pasture for $55.50. A put in 5 cows, 12 
weeks; B ; 4 cows, 10 weeks; and C, 6 cows, 8 weeks; how much 
ought each to pay? Ans. A $22.50; B $15; C $18. 

3. B commenced business with a capital of $15000. Three 
months afterward C entered into partnership with him, and put 
in 125 acres of land. At the close of the year their profits were 
$4500, of which C was entitled to $1800 ; what was the value of 
the land per acre ? 

4. A and B engaged in trade. A put in $4200 at first, and 9 
months afterward $200 more. B put in at first $1500, and at the 
end of 6 months took out $500. At the end of 16 months their 
gain was $772.20 ; how much is the share of each ? 

5. Four companies of men worked on a railroad. In the first 
company there were 30 men who worked 12 days, 9 hours a day; 
in the second, there were 32 men who worked 15 days, 10 hours 



PARTNERSHIP. 



365 



a day; in the third, there were 28 men who worked 18 days, 11 
hours a day; and in the fourth, there were 20 men who worked 
15 days, 12 hours a day. The entire amount paid to all the com- 
panies was $1500; how much were the wages of each company? 

6. A and B are partners. A's capital is to B's as 5 to 8 ; at 
the end of 4 months A withdraws J of his capital, and B f of his ; 
at the end of the year their whole gain is $4000 ; how much be- 
longs to each ? Ans. A, $1714§ ; B, $2285$.' 

7. B, C, and D form a manufacturing company, with capitals 
of $15800, $25000, and $30000 respectively. After 4 months B 
draws out $1200, and in 2 months more he draws out $1500 more, 
and 4 months afterward puts in $1000. C draws out $2000 at 
the end of 6 months, and $1500 more 4 months afterward, and a 
month later puts in $800. D puts in $1800 at the end of 7 
months, and 3 months after draws out $5000. If their gain at 
the end of 18 months be $15000, how much should each receive? 

Ans. B, $2133.82; C, $5104.62; D, $6761.56. 

8. The joint stock of a company was $5400, which was doubled 
at the end of the year. A put i for f of a year, B § for J a year, 
and C the remainder for one year. How much is each one's share 
of the entire stock at the end of the year ? 

9. Three men engage in merchandising. A's money was in 
10 months, for which he received $456 of the profits ; B's was in 
8 months, for which he received $343.20 of the profits; and C's 
was in 12 months, for which he received $750 of the profits. Their 
whole capital invested was $14345 ; how much was the capital of 
each? Ans. A's, $4332; B's, $7075.50; C\s, $5937.50. 

10. Three men take an interest in a coal mine. B invests his 
capital for 4 months, and claims T ^ of the profits ; C's capital is in 
.8 months ; and D invests $6000 for 6 months, and claims § of the 
profits ; how much did B and C put in ? 

11. A, B, and C engage in manufacturing shoes. A puts in 
$1920 for 6 months ; B, a sum not specified for 12 months ; and 
C, $1280 for a time not specified. A received $2400 for his stock 
and profits, B $4800 for his, and C $2080 for his. Bequired, 
B's stock, and C's time ? 

31* 



366 ALLIGATION. 



ALLIGATION. 

631. Alligation treats of mixing or compounding two or 
more ingredients of different values or qualities. 

632. The Mean Price or Quality is the average price or 
quality of the ingredients, or the price or quality of a unit of the 
mixture. 

CASE I. 

633. To find the mean price or quality of a mixture, 
when the quantity and price of the several ingredient* 
are given. 

Note. — The process of finding the mean or average price of several ingredi- 
ents is called Alligation Medial. 

1. A produce dealer mixed together 84 bushels of oats worth 
$.30 a bushel, 60 bushels of oats worth $.38 a bushel, and 56 
bushels of oats worth $. 40 a bushel ; required, the mean price. 

operation. Analysis. The worth of 84 

$.30 X 84 = $25.20 bushels @ $.30 is $25.20, of 

.38 X 60 = 22.80 60 bushels @ $.38 is $22.80, 

.40 X 56 = 22.40 and of 56 bushels @ $.40 is 

200 ^ $70 40 $22.40 ; and we have in the 

J whole compound 84 + 60 + 56 

$.3520, Ans. =200 bushels, worth $25.20+ 

$22.80 + $22.40 = $70.40. One bushel of the mixture is therefore 
worth $70.40 -T- 200 = $.352. Hence the following 

Rule. Find the entire cost or value of the ingredients, and 
divide it by the sum of the simples. 

EXAMPLES FOR PRACTICE. 

1. A grocer mixed 4 lb. of tea at $.60 with 3 lb. at $.70, 1 lb. 
at $1.10, and 2 lb. at $1.20; how much is 1 lb. of the mixture 
worth? Ans. $.80. 

2. A dealer in liquors would mix 14 gal. of water with 12 gal. 
of wine at $.75, 24 gal. at $.90, and 16 gal. at $1.10; how much 
is a gallon of the mixture worth ? Ans. $.73 5 1 3- 



ALLIGATION. 367 

3. If 3 lb. 6 oz. of gold 23 carats fine be compounded with 
4 lb. 8 oz. 21 carats, 3 lb. 9 oz. 20 carats, and 2 lb. 2 oz. of alloy, 
what is the fineness of the composition ? Ans. 18 carats. 

4. A grain dealer mixes 15 bu. of wheat at $1.20 with 5 bu. 
at §1.10, 5 bu. at $.90, and 10 bu. at $70; what will be his gain 
per bushel if he sell the compound at $1.25. 

5. A merchant sold 17 lb. of sugar at 5 cts. a pound, 51 lb. at 
8 cts., 68 lb. at 10 cts., 17 lb. at 12 cts., and thereby gained on 
the whole 33 J per cent. ; how much was the average cost per 
pound ? 

6. A drover bought 42 sheep at $2.70 per head, 48 at $2.85, 
and 65 at $3.24; at what average price per head must he sell 
them to gain 20 per cent.? Ans. $3,588. 

7. A surveyor took 10 sets of observations with an instrument, 
for the measurement of an angle, with the following results : 1st, 
36° 17' 25.4"; 2d, 36° 17' 24.5"; 3d, 56° 17' 27.8"; 4th, 36° 17' 
26.9"; 5th, 36° 17' 25.4"; 6th, 36° 17' 24.7"; 7th, 36° 17' 24.2"; 
8th, 36° 17' 26.3";. 9th, 36° 17' 25.8"; 10th, 36° 17' 26.7". What 
is the average of these measurements ? Ans. 36° 17' 25.8". 

8. Three trials were made with chronometers to determine the 
difference of time between two places; the first trial gave 37 min. 
54.16 sec, the second 37 min. 55.56 sec, and the third 37 min. 
54.82 sec. Owing to the favorable conditions of the third trial, 
it is entitled to twice the degree of reliance to be placed upon 
either of the others ; what should be taken as the difference of 
longitude between the two places, according to these observations ? 

Ans. 9° 28' 22.46". 

CASE II. 

634. To find the proportional quantity to be used of 
each ingredient, when the mean price and the prices of 
the several simples are given. 

Note. — The process of finding the quantities to be used in any required mix- 
ture is commonly called Alligation Alternate. 

1. A farmer would mix oats worth 3 shillings a bushel with 
peas worth 8 shillings a bushel, to make a compound worth 5 shil- 
lings a bushel ; what quantities of each may he take ? 



368 



ALLIGATION. 



OPERATION. 

3 | * | 3 

*|2 



f8|*|8) 

|8 I * I 2 J 



Ans. 



Analysis. If a mixture, in any pro- 
portions, of oats worth 3 shillings a 
bushel and peas worth 8 shillings, be 
priced at 5 shillings, there will be a 
gain on the oats, the ingredient worth less than the mean price, and 
a loss on the peas, the ingredient worth more than the mean price ; 
and if we take such quantities of each that the gain and loss shall 
each be 1 shilling, the unit of value, the result will be the required 
mixture. By selling 1 bushel of oats worth 3 shillings for 5 shil- 
lings, there will be a gain of 5 — 3 = 2 shillings, and to gain 1 shil- 
ling would require J of a bushel ; hence we place J opposite the 3. By 
selling 1 bushel of peas worth 8 shillings for 5 shillings, there will 
be a loss of 8 — 5 = 3 shillings, and to lose 1 shilling will require £ 
of a bushel ; hence we write i opposite the 8. Therefore, J bushel of 
oats to J of a bushel of peas are the proportional quantities for the 
required mixture. It is evident that the gain and loss will be equal, 
if we take any number of times these proportional terms for the mix- 
ture. We may therefore multiply the fractions i and J by 6, the least 
common multiple of their denominators, and obtain the integers 3 
and 2 for the proportional terms (418,111); that is, we may take, for the 
mixture, 3 bushels of oats to 2 bushels of peas. 

2. What relative quantities of sugar at 7 cents, 8 cents, 11 cents, 
and 14 cents per pound, will produce a mixture worth 10 cents 
per pound? 

operation. Analysis. To preserve the 

equality of gains and losses, we 
must compare two prices or sim- 
ples, one greater and one less than 
the mean rate, and treat each pair 
or couplet as a separate example. 
Thus, comparing the simples whose 
prices are 7 cents and 14 cents, we 
find that, to gain 1 cent, J of a 
pound at 7 cents must be taken, 
and, to lose 1 cent, J of a pound at 
14 cents must be taken ; and com- 
paring the simples the prices of 
which are 8 cents and 11 cents, we 
find that £ pound at 8 cents must be taken to gain 1 cent, and 1 pound 
at 11 cents must be taken to lose 1 cent. These proportional terms are 




1 


2 


3 


4 


5 


T 




4 




4 




i 




1 


1 




1 




2 


2 


i 




3 




3 



.Or, 



10 1 





1 


2 


3 


( 7 


T 




T 


8 




i 


4 


11 


i 




3 


14 




I 


2 



ALLIGATION. 309 

-written in columns 1 and 2 We now reduce these couplets separately 
to integers, as in the last example, writing the results in columns 3 
and 4 ; and arranging all the terms in column 5, we have 4, 1, 2, and 
3 for the proportional quantities required. If we compare the prices 
7 and 11 for the first couplet, and the prices 8 and 14 for the second 
couplet, as in the second operation, we shall obtain 1, 4, 3, and 2 fof 
the proportional terms. 

It will be seen that in comparing the simples of any couplet, one 
of which is greater and the other less than the mean rate, the pro- 
portional number finally obtained for either term is the difference 
between the mean rate and the other term. Thus, in comparing 7 
and 14, the proportional number corresponding to the former simple 
is 4, which is the difference between 7 and the mean rate 10 ; and 
the proportional number corresponding to the latter simple is 3, 
which is the difference between 7 and the mean rate. The same is 
true of every other couplet. Hence, when the simples and the mean 
rate are integers, the intermediate steps taken to obtain the final pro- 
portional numbers as in columns 1, 2, 3, and 4, may be omitted, and 
the same results readily found by taking the difference between each 
simple and the mean rate, and placing it opposite the one with which 
it is compared. 

From these examples and analyses we derive the following 

Eule. I. Write the prices or qualities of the several ingre- 
dients in a column, and the mean price or quality at the left. 

II. Consider any two prices, one of which is less and the other 
greater than the mean rate, as forming a couplet ; find the differ- 
ence between each of these prices and the mean rate, and write the 
reciprocal of each difference opposite the other term in the couplet, 
as one of the proportional terms. In like manner form the couplets, 
till all the prices have been employed, writing each pair of propor- 
tional terms in a separate column. 

. III. If the proportional terms thus obtained are fractional, mul- 
tiply each pair by the least common multiple of their denominators, 
and carry these integral products to a single column, observing to 
add any two or more that stand in the same horizontal line ; the 
final results voill be the 'proportional quantities required. 

Notes. — 1. If the numbers in any couplet or column have a common factor, 
it may be rejected. 





OPERATION. 




5i 


4 
3 


1 


4 




4 


6 




4j 




12 


12 


71 


1 


tl 


2 


2 


4 



370 ALLIGATION. 

2. We may also multiply the numbers in any couplet or column by any mul- 
tiplier we choose, without affecting the equality of the gains and losses, and 
thus obtain an indefinite number of results, any one of which being taken will 
give a correct final result. 

EXAMPLES FOR PRACTICE. 

1. What quantities of flour worth $5J, $6, and $7f per barrel, 
must be sold, to realize an average price of $6i per barrel ? 

Analysis. Comparing the 
first price with the third, we ob- 
6 J \ 6 4 j 12 12 *ain the couplet f to f ; and com- 

paring the second price with the 
third, we obtain the couplet 4 to 
§ . Reducing these proportional terms to integers, we find that we 
may take 4 barrels of the first kind with 2 of the third, and 12 of the 
second kind with 2 of the third ; and these two combinations taken 
together give 4 of the first kind, 12 of the second, and 4 of the third. 

2. How much sugar worth 5 cts., 7 cts., 12 cts,, and 13 cts. per 

pound, will form a mixture worth 10 cts. per pound ? 

A f 3 lb. of each of the first and third kinds, 2 lb. 
Ans. \ ' 

I of the second, and 5 lb. of the fourth. 

3. How can wine worth $60, $90, and $1.15 per gallon be mixed 
with water so as to form a mixture worth $.75 a gallon ? 

j (By taking 3 gal. of each of the first two kinds of 
1 wine, 15 gal. of the third, and 8 gal. of water. 

4. A farmer has 3 pieces of land worth $40, $60, and $80 an 
acre respectively How many acres must he sell from the dif- 
ferent tracts, to realize an average price of $62.50 an acre ? 

5. How much wine worth $.60, $.50, $.42, $.38, and $.30 per 
pint, will make a mixture worth $.45 a pint ? 

6. What relative quantities of silver f pure, | pure, and T 9 ^ 
pure, will make a mixture f pure ? 

Ans. 1 lb. | pure, 5 lb. | pure, and 3 lb. T % pure. 

CASE III. 

635. When two or more of the quantities are re- 
quired to be in a certain proportion. 

1. A farmer having oats worth $.30, corn worth $.60, and wheat 



ALLIGATION. 



371 



r 30 
50 ] 60 

(no 



1 


2 


3 


4 


5 


6 


1 

25 


i 

20 


1 


3 


6 


7 


1 
10 




2 






2 




1 

SOi 




1 


2 


2 



worth $1.10 per bushel, desires to form a mixture worth $.50 per 
bushel, which shall contain equal parts of corn and wheat; in 
what proportion shall the ingredients be taken ? 

operation. Analysis. We first obtain 

the proportional terms in col- 
umns 3 and 4, by Case II. 
Now, it is evident that the loss 
and gain will be equal if we 
take each couplet, or any mul- 
tiple of each, alone; or both 
couplets, or any multiples of both, together. Multiplying the terms 
in column 4 by 2, we obtain the terms in column 5 ; and adding the 
terms in columns 3 and 5, we obtain the terms in column 6 ; that is, 
the farmer takes 7 bushels of oats to 2 of corn and 2 of wheat, which 
is the required proportion. Hence the following 

Rule. I. Compare the given prices, and obtain the proportional 
terms by couplets, as in Case IL 

II. Reduce the couplets to higher or lower terms, as may be re- 
quired; then select the columns at pleasure, and combine them by 
adding the terms in the same horizontal lin*e, till a set of pro- 
portional terms is obtained, answering the required conditions, 

EXAMPLES FOR PRACTICE. 

1. A grocer has four kinds of molasses, worth $.25, $.50, $.62, 
and $-70 per gallon, respectively ; in what proportions may he mix 
the four kinds, to obtain a compound worth $.58 per gallon, using 
equal parts of the first two kinds ? Ans, 4, 4, 8 and 11. 

2. In what proportions may we take sugars at 7 cts., 8 cts., 13 
cts., and 15 cts., to form a compound worth 10 cts. per pound, using 
equal parts of the first three kinds ? Ans. 5, 5, 5 and 2. 

3 A miller has oats at 30 cts., corn at 50 cts., and wheat at 

100 cts, per bushel. He desires to form two mixtures, each worth 

70 cts. per bushel. In the first he would have equal parts of oats 

and corn, and in the second, equal parts of corn and wheat; what 

must be the proportional terms for each mixture ? 

A ( For the first mixture, 1, 1 and 2. 
Ans, J 



1 For the second mixture, 1, 4 and 4. 



372 



ALLiaATION. 



CASE IV. 

636. When the quantity of one of the simples is 
limited. 

1. A miller has oats worth $.28, corn worth $.44, and barley 
worth $.90 per bushel. He wishes to form a mixture worth $.58 
per bushel, and containing 100 bushels of corn. How many 
bushels of oats and barley may he take ? 

OPERATION. 



58 5 
(.93JJ 



28 j 

44]! 



ft 




7 




7 


140 




1 
14 




5 


5 


100 


1 

'3^ 


ft 


6 


2 


8 


160 



Analysis. By Case 
II, we find the pro- 
portional quantities 
to be 7 bushels of 
oats to 5 of corn and 
8 of barley. But as 100 bushels of corn, instead of 5, are required, 
we must take l ° 5 ° =20 times each of the other ingredients, in order 
that the gain and loss may be equal ; and we shall therefore have 
7 X 20 = 140 bushels of oats, and 8 X 20 = 160 bushels of barley. 
Hence the following 

Rule. Find the proportional quantities by Case II or Case 
III. Divide the* given quantity by the proportional quantity of 
this ingredient, and multiply each of the other proportional quan- 
tities by the quotient thus obtained. 



EXAMPLES FOR PRACTICE. 

1. A dairyman bought 10 cows at $20 a head ; how many must 
he buy at $16, $18, and $24 a head, so that the whole may cost 
him an average price of $22 a head ? 

Arts. 10 at $16, 10 at $18, and 30 at $24. 

2. Bought 12 yards of cloth for $15 ; how many yards must I 
buy at $lf , and $J a yard, that the average price of the whole 
may be $1^? Ans. 36 yards at $lf and 52 yards at $f. 

3. How much water will dilute 9 gal. 2 qt. 1 pt. of alcohol 96 
per cent, strong to 84 per cent. ? Ans. 1 gal. 1 qt. 1 pt. 

4. A grocer mixed teas worth $.30, $.55, and $.70 per pound 
respectively, forming a mixture worth $.55 per pound, having equal 
parts of the first two kinds, and 12 lbs. of the third kind ; how 
many pounds of each of the first two kinds did he take ? 



ALLIGATION. 373 

CASE V. 

637. When the quantities of two or more of the in- 
gredients are limited. 

1. How many bushels of rye at $1.08, and of wheat at $1.44, 
must be mixed with 18 bushels of oats at $.48, 8 bushels of corn 
at $52, and 4 bushels of barley at $.85, that the mixture may be 
worth $.84 per bushel ? • 

OPERATION. ' A *ALYSIS. Of the given 

8 48* 18== 8 8 64 quantities there are 18 -f- 

'f-n o 4*1 fi 8 + 4 = 30 bushels, whose 



.85 x 4 = 3.40 



mean or average price we 
find by Case I to be $.54. 



30 ) $16.20 "VVe are therefore required to 

Mean price of the \ * r* mix 30 bushels of grain 

given simples j worth $.54 per bushel, with 

, 54 _i_ I _i^ 1 4 J 2 I 6 i 30 rye at $1.08, and wheat at 

84 \ 108 M 5 1 5 25 $1-44, to make a compound 



JL 
30 


1 
30" 


4 


2 


6 


30 


1 
24 




5 




5 


25 




1 
50 




1 


1 


5 



/ 144 ' i_ I 1 1 1 1 1 5 worth $.84 per bushel. Pro- 

ceeding as in Case IV, we 
find there will be required 25 bushels of rye, and 5 bushels of wheat. 
Hence the following 

Rule. Consider those ingredients whose quantities and prices 
are given as forming a mixture, and find their mean price by Case 
I; then consider this mixture as a single ingredient ivhose quantity 
and price are knoicn, and find the quantities of the other ingredients 

by Case IV. 

EXAMPLES EOR PRACTICE. 

1. A gentleman bought 7 yards of cloth @ $2.20, and 7 yards 
@ $2 ; how much must he buy @ $1.60, and @ $1.75 that the 
average price of the whole may be $1.80 ? 

2. How much wine, at $1.75 a gallon, must be added to 60 gal- 
lons at $1.14, and 30 gallons at $1.26 a gallon, so that the mixture 
may be worth $1.57 a gallon ? Ans. 195 gallons. 

3. A farmer has 40 bushels of wheat worth $2 a bushel, and 
70 bushels of corn worth $} a bushel. How many oats worth $} 
a bushel must he mix with the wheat and corn, to make the mix- 
ture worth $1 a bushel ? Ans. 6f bushels. 

32 



374 



ALLIGATION. 



CASE VI. 

638. When the quantity of the whole compound is 
limited. 

1. A tradesman has three kinds of tea rated at $.30, $.45, and 
$.60 per pound, respectively; what quantities of each should he 
take to form a mixture of 72 pounds, worth $.40 per pound ? 

operation. ' Analysis. By Case II, 

we find the proportional 

quantities to form the 

mixture to be 8 lb. at 

40 ■} 45 T V 2 2 18 $.30, 2 lb. at $.45, and 

1 lb. at $.60. Adding 
these proportional quanti- 
ties, we find that they 
would form a mixture of 8 pounds. And since the required mixture 
is y = 9 times 8 pounds, we multiply each of the proportional terms 
by 9, and obtain for the required quantities, 27 lb. at $.30, 18 lb. at 
$.45, and 9 lb. at $.60. Hence the following 

Rule. Find the 'proportional numbers as in Case II or Case 
III Divide the given quantity by the sum of the proportional 
quantities, and multiply each of the, proportional quantities by the 
quotient thus obtained. 



1 


2 


3 


4 


5 


6 


1 

10 


i 

10 


2 


3 


5 


45 




i 

1 5 




2 


2 


18 


1 

20 




1 




1 
~6 


9 
54 



EXAMPLES FOR PRACTICE. 

1. A grocer has coffee worth 8 cts., 16 cts., and 24 cts. per 
pound respectively ; how much of each kind must he use, to fill a 
cask holding 240 lb, that shall be worth 20 cts. a pound ? 

Ans. 40 lb. at 8 cts., 40 lb. at 16 cts., and 160 lb. at 24 cts. 

2. A man bought calves, sheep, and lambs, 154 in all, for $154. 
He paid $3 J for each calf, $1J for each sheep, and $| for each 
lamb ; how many did he buy of each kind ? 

Ans. 14 calves, 42 sheep, and 98 lambs. 

3. A man paid $165 to 55 laborers, consisting of men, women, 
and boys ; to the men he paid $5 a week, to the women $1 a week, 
and to the boys $ J a week ; how many were there of each ? 

Ans. 30 men, 5 women, and 20 boys. 



involution. 375 



INVOLUTION. 

($09* A Power is the product arising "from multiplying a 
number by itself, or repeating it any number of times as a factor. 

640- Involution is the process of raising a number to a given 
power. 

©41 * The Square of a number is its second power. 

043 The Cfcbe of a number is its third power. 

043. In the process of involution, we observe, 

I. That the exponent of any power is equal to the number of 
times the root has been taken as a factor in continued multiplica- 
tion. Hence 

II. The product of any two or more powers of the same num- 
ber is the power denoted by the sum of their exponents, and 

III. If any power of a number be raised to any given power, 
the result will be that power of the number denoted by the pro- 
duct of the exponents. 

1. What is tfce 5th power of 6 ? 

operation. Analysis. We 

6x6x6x6x6 = 7776, Ans. multiply 6 by it- 

Or se lf> an d this pro- 

6x6 = 6 2 = 86 ^ uct ky 6> an d so 

36 X 6 = 6 3 = 216 on > un til 6 has 

6 3 X 6 7 = 6 5 = 216 X 36 = 7776, A?is. beentaken5times 

in continued niuk 
tiplication ; the final product, 7776, is the power required, (I). Or, 
we may first form the 2d and 3d powers • then the product of these 
two powers will be the 5th power required, (II). 

2. What is the 6th power of 12 ? 

operation. Analysis. We find the cube of 

% ■"--■ == m „„ t . the second power, which must be 

144' = 2985984, Ans. the 6th ^ (m) _ 

644* Hence for the involution of numbers we have the fol- 
lowing 



376 INVOLUTION. 

Rule. I. Multiply the given number by itself in continued 
multiplication, till it has been takeu as many times as a factor as 
there are units in the exponent of the required power. Or, 

II. Multiply together two or more powers of the given number, 
the sum of whose exponents is equal to the exponent of the required 
power. Or, 

III Raise some power of the given number to such a power 

that the product of the two exponents shall be equal to the exponent 

of the required power. 

Notes. — 1. A fraction is involved to any power by involving each of its 
terms separately to the required power. 

2. Mixed numbers should be reduced to improper fractions before involution. 

3. When the number to be involved is a decimal, contracted multiplication 
may be applied with great advantage. 

EXAMPLES FOR PRACTICE. 

1. What is the square of 79 ? Ans 6241. 

2. What is the cube of 25.4 ? Ans. 16387.064. 

3. What is the square of 1450 ? 

4. Raise 16| to the 4th power. Ans. 79659|f| 

5. Raise 2 to the 20th power. Ans. 1048576 

6. Raise .4378565 to the 8th power, reserving 5 decimals. 

Ans. .00135 4z 
- 7. Raise 1.052578 to the 6th power, reserving 4 decimals 

Ans. 1.3600 zh. 

8. Involve .029 to the 5th power? 

Ans. .000000020511149. 
Find the value of each of the following expressions : 

9. 4.367*. Ans. 363.691179. 
10 (!)•. Ans. ff|. 
11. (2f)*. Ans. 157 T W_. 
12 4.6 3 x 25 3 Ans. 15646 16. 
13. (6f )* — 7.25*. 14. (Sif x 2.5 2 

15. | of (|) 3 of (34) 2 . Ans. 5f. 

Note.— Cancel like powers of the same factor. 

16. 7 8 +-3 08 5 . 

17 (4' x 5 6 X 12*) h- (4 2 x 10 1 X 3 2 ). Ans. 2. 



EVOLUTION. 377 

EVOLUTION. 

64t>. A Root is a factor repeated to produce a power ; thus, 
in the expression 7x7x7 = 343, 7 is the root from which the 
power, 343, is produced. 

646. Evolution is the process of extracting the root of a 
number considered as a power ; it is the reverse of Involution. 

Any number whatever may be considered a power whose root 
is to be extracted. 

©4:7. A Rational Root is a root that can be exactly obtained. 

©48. A Surd is an indicated root that can not be exactly ob- 
tained. 

©49. The Radical Sign is the character, </, which, placed 
before a number, indicates that its root is to be extracted. 

©cSO. The Index of the root is the figure placed above the 
radical sign, to denote what root is to be taken. When no index 
is written, the index, 2, is always understood. 

©31. The names of roots are derived from the corresponding 
powers, and are denoted by the indices of the radical sign. Thus, 
\/l00 denotes the square root of 100; VlOO denotes the cube 
root of 100; V^lOO denotes the fourth root of 100; etc. 

©52. Evolution is sometimes denoted by a fractional exponent, 
the name of the root to be extracted being indicated by the deno- 

minator. Thus, the square root of 10 may be written 10 2 ; the 
cube root of 10, 10 , etc. 

©S3. Fractional exponents are also used to denote both invo*- 
lution and evolution in the same expression, the numerator indi- 
cating the power to which the given number is to be raised, and 
the denominator the root of the power which is to be taken ; thus, 

7 denotes the cube root of the second power of 7, and is the 

same as v 7 2 ; so also 7 2 = v 7 5 . 

©34. In extracting any root of a number, any figure or figures 

may be regarded as tens of the next inferior order. Thus, in 

2546, the 2 may be considered as tens of the 3d order, the 25 as 

tens of the second order, or the 254 as tens of the first order. 
32* 



378 EVOLUTION. 

SQUARE ROOT. 

G55* The Square Root of a number is one of the two equal 
factors that produce the number. Thus, the square root of 64 is 
8, for 8 x 8 = 64. 

To derive the method of extracting the square root of a num- 
ber, it is necessary to determine 

1st. The relative number of places in a number and its square root. 

2d. The relations of the figures of the root to the periods of 
the number. 

3d. The law by which the parts of a number are combined in 
the formation of its square ; and 

4th. The factors of the combinations. 

C£S6. The relative number of places in a given number and 
its square root is shown in the following illustrations. 

Roots. Squares. Roots. Squares. 

1 1 1 1 

9 81 10 1,00 

99 98,01 100 1,00,00 

999 99,80,01 1000 1,00,00,00 

From these examples we perceive 

1st. That a root consisting of 1 place may have 1 or 2 places in the 
square. 

2d. That in all cases the addition of 1 place to the root adds 2 
places to the square. Hence, 

I. If we point off a number into two-figure periods, commencing 
at the right hand, the number of full periods and the left hand 
full or partial period will indicate the number of places in the 
square root 

To ascertain the relations of the several figures of the root to the 
periods of the number, observe that if any number, as 2345, be de- 
composed at pleasure, the squares of the left hand parts will be re 
lated in local value as follow3 : 

2000 2 = 4 00 00 00. 
2300 2 = 5 29 00 00 
2340 2 = 5 47 56 00 
2345" 2 = 5 49 90 25 : Hence, 
IT. The square of the first figure of the root is contained wholly 
in the first period of the power ; the square of the first two figures 



SQUARE ROOT. 



379 



of the root is contained wholly in the first two periods of the power ; 
and so on. 

Note. — The periods and figures of the root are counted from the left hand. 

The combinations in the formation of a square may be shown as 
follows : 

If we take any number consisting of two figures, as 43, and decom- 
pose it into two parts, 40 + 3, then the square of the number may 
be formed by multiplying both parts by each part separately : thus, 

40 + 3 
40 + 3 

120 + 9 
160C + 120 

43* = 1600 + 240 + 9 = 1849. 
Of these combinations, we observe that the first, 1600, is the square 
of 40 the second, 240, is twice 40 multiplied by 3 ; and the third, 9, 
is the square of 3. Hence, 

III. The square of a number composed of tens and units is 
equal to the square of the tens, plus twice the tens multiplied by the 
units, plus the square of the units. 

By observing the manner in which the square is formed, we per- 
ceive that the unit figure must always be contained as a factor in 
both the second and third parts ; these parts taken together, may 
therefore be factored, thus, 240 + 9 = (80 + 3) X 3. Hence, 

IV. If the square of the tens be subtracted from the entire 
square, the remainder will be equal to twice the tens plus the units 
multiplied by the units. 

1. What is the square root of 5405778576 ? 

Analysis. Pointing off the 
( 73524 given number into periods of 
two figures each, the 5 periods 
show that there will be 5 fig- 
ures in the root, (I). Since 
the square of the first figure 
of the root is always contained 
wholly in the first period of 
the power, (II), we seek for the 
greatest square in the first pe~ 
riod, 54, which we find by 
trial to be 49, and we place 



143 


OPERATION. 

5405778576 
49 

505 
429 


1465 


7677 
7325 


14702 


35285 
29404 


147044 


588176 
588116 



380 EVOLUTION. 

its root, 7, as the first figure of the required root, and regard it as 
tens of the next inferior order, (II). We now subtract 49, the 
square of the first figure of the root, from the first period, 54, and 
bringing down the next period, obtain 505 for a remainder. And 
since the square of the first two figures of the root is contained wholly 
in the first two periods of the power, (II), the remainder, 505, must 
contain at least twice the first figure (tens) plus the second figure 
(units), multiplied by the second figure, (IV). Now if we could divide 
this remainder by twice the first figure plus the second, which is one 
of the factors, the quotient would be the second figure, or the other 
factor. But since we have not yet obtained the second figure, the 
complete divisor can not now be employed ; and we therefore write 
twice the first figure, or 14, at the left of 505 for a trial divisor, re- 
garding it as tens. Dividing the dividend, exclusive of the right 
hand figure, by 14, we obtain 3 for the second, or trial figure of the 
root, which we annex to the trial divisor, 14, making 143, the com- 
plete divisor. Multiplying the complete divisor by the trial figure 
3, and subtracting the product from the dividend, we have 7 6 for a 
remainder. We have now taken the square of the first two figures of 
the root from the first two periods ; and since the square of the first 
three figures of the root is contained wholly in the first three periods, 
(II), we bring down the third period, 77. to the remainder, 76, and 
obtain for a new dividend 7677, which must contain at least twice the 
two figures already found plus the third, multiplied by the third, (IV). 
Therefore to obtain the third figure, we must take for a new trial 
divisor twice the two figures, 73, considered as tens of the next infe- 
rior order, which we obtain in the operation by doubling the last fig- 
ure of the last complete divisor, 143, making 146. Dividing, we ob- 
tain 5 for the next figure of the root ; then regarding 735 as tens of 
the next inferior order, we proceed as in the former steps, and thus 
continue till the entire root, 73524, is obtained. 

6o"S7. From these principles and illustrations we derive the 
following 

Rule. I. Point off the given number into periods of two figures 
each, counting from units' place toward the left and right. 

II. Find the greatest square number in the left hand period, and 
write its root for the first figure in the root ; subtract the square 
number from the left hand 'period, and to the remainder bring 
down the next period for a dividend. 



SQUARE ROOT. ggj 

III. At the left of the dividend write twice the first figure of the 
root, for a trial divisor ; divide the dividend, exclusive of its right 
hand figure, by the trial divisor, and write the quotient for a trial 
figure in the root. 

1\ . Annex the trial figure of the root to the trial divisor for a 
complete divisor ; multiply the complete divisor by the trial figure 
in the root, subtract the product from the dividend, and to the 
remainder bring down the next period for a new dividend. 

If. Multiply the last figure of the last complete divisor by 2 for 
a new trial divisor with which proceed as before. 

Notes. — 1. If at any time the product be greater than the dividend, diminish 
the trial figure of the root, and correct the erroneous work. 

2. If a cipher occur in the root, annex a cipher to the trial divisor, and another 
period to the dividend, and proceed as before. 

3. If there is a remainder after all the periods have been brought down, 
annex periods oi ciphers, and continue the root to as many decimal places as 
are required. 

4. The decimal points in the work may be omitted, care being taken to point 
off in the root according to the number of decimal periods used. 

5. The square root of a common fraction may be obtained by extracting the 
square roots of the numerator and denominator separately, provided the terms 
are perfect squares ; otherwise, the fraction may first be reduced to a decimal. 

6. Mixed numbers may be reduced to the decimal form before extracting the 
root ; or, if the denominator of the fraction is a perfect square, to an improper 
fraction. 

7. The pupil will acquire greater facility, and secure greater accuracy, by 
keeping units of like order under each other, and each divisor opposite the 
corresponding dividend, as shown in the operation. 

EXAMPLES FOR PRACTICE. 

1. What is the sqnare root of 31584.4 ? Ans. 562. 

2. What is the square root of 152399025 ? Ans. 12345. 

3. What is the square root of 56280004 ? Of 597 ? ; 

4. What is the square root of 10795.21 ? Ans. 103.9. 

5. What is the square root of 58.1406J ? Ans. 7.62f . 
Find the values of the following expressions : 

■: 6. ^/ .0000316969. Ans. .00563. 

7. v / 3858.07694409"64. Ans. 62.11342. 

8. v"|. Ans. .745355+. 



9. </99225 — 63504. 10. V.126736— \/.045369 

11. •jp_X ^flfl Ans. I 

12. V81 2 x 625 2 x 2*. Ans. 900. 



13 
1 6' 



48 


400 

384 


562 


1600 
1124 


5648 


47600 
45184 


5656 


2416* 

2262 


566 


154 
113 



382 EVOLUTION. 

CONTRACTED METHOD. 

658. 1. Find the square root of 8, correct to 6 decimal 
places. 

operation. Analysis. Extracting the square 

12.828427+ , Ans. root in the usual way until we have 
8 000000 obtained the 4 places, 2.828, the 

4 corresponding remainder is 2416, and 

the next trial divisor, with the cipher 
omitted, is 5656. We now omit to 
bring down a period of ciphers to 
the remainder, thus contracting the 
dividend 2 places ; and we contract 
the divisor an equal number of places 
by omitting to annex the trial figure 
of the root, and regarding the right 
hand figure, 6, as a rejected or re- 
dundant figure. We now divide as 
- 7 in contracted division of decimals, 

aq (226), bringing down each divisor in 

its place, with one redundant figure 
increased by 1 when the rejected figure is 5 or more, and carrying the 
tens from the redundant figure in multiplication. We observe that 
the entire root, 2. 828427+? contains as many places as there are places 
in the periods used. Hence the following 

Rule. L If necessary, annex periods of ciphers to the given 
number j and assume as many figures as there are places required 
in the root ; then proceed in the usual manner until all the assumed 
figures have been employed, omitting the remaining figures } if any. 
II. Form the next trial divisor as usual, but omit to annex to it 
the trial figure of the root, reject one figure from the right to form 
each subsequent divisor, and in multiplying regard the right hand 
figure of each contracted divisor as redundant. 

Notes. — 1. If the rejected figure is 5 or more, increase the next left hand 
figure by 1, 

2. Always take full periods, both of decimals and integers. 

EXAMPLES FOR PRACTICE. 

1. Find the square root of 32 correct to the seventh decimal 
place. Ans. 5.6568542+. 



cube root. 383 

2. Find the square root of 12 correct to the seventh decimal 
place. Arts. 3.4641616 + . 

3. Find the square root of 3286.9835 correct to the fourth 
decimal place. Ans. 57.3330+. 

4 Find the square root of .5 correct to the sixth decimal 
place. Ans. .745355 + . 

5. Find the square root of 6| correct to the sixth decimal 
place. Ans. 2.563479 + . 

6. Find the square root of 1.06 5 correct to the sixth decimal 
place. Ans. 1.156817+ . 

3 

7. Find the value of 1.0125^ correct to the fourth decimal 
place. Ans. 1.0187+. 

8. Find the value of 1.023375^ correct to the sixth decimal 
place. Ans. 1.083586+ . 

CUBE ROOT. 

Go9. The Cube Root of a number is one of the three equal 
factors that produce the number. Thus, the cube root of 343 is 
7, since 7x7x7 = 343. 

To derive the method of extracting the cube root of a number, 
it is necessary to determine 

1st. The relative number of places in a given number and its 
cube root. 

2d. The relations of the figures of the root to the periods of 
the number. 

3d. The law by which the parts of a number are combined in 
the formation of a cube ; and 

4th. The factors of these combinations. 

OGOe The relative number of places in a given number and 
its cube, is shown in the following illustrations : 



Roots. 


Cubes. 


Roots. 


Cubes. 


1 

9 

99 

999 


1 

729 

907,299 

997,002,999 


1 

10 

100 

1000 


1 

1,000 

1,000,000 

1,000,000,000 



From these examples, we perceive, 



384 EVOLUTION. 

1st. That a root consisting of 1 place may have from 1 to 3 places 
in the cube. 

2d. That in all cases the addition of 1 place to the root adds 3 
places to the cube Hence, 

I. If we point off a number into three-figure periods, com- 
mencing at the right hand, the number of full periods and the left 
hand full or partial period will indicate the number of places in 
the cube root. 

To ascertain the relations of the several figures of the root to the 
periods ol the number, observe that if any number, as 5423, be de- 
composed, the cubes of the parts will be related in local value, as 
follows : 

5000 3 = 125 000 000 000 
5400 3 = 157 364 000 000 
5420 3 ■-= 159 220 088 000 
5423 3 == 159 484 621 967. Hence, 

II The cube of the first figure of the root is contained wholly in 
the first period of the power ; the cube of the first two figures of 
the root is contained wholly in the first two periods of the power ; 
and so on 

To learn the combinations of tens and units in the formation of a 
cube, take any number consisting of two figures, as 54, and decom- 
pose it into two parts, 50+4 ; then having formed the square by 656, 
III, multiply each part of this square by the units and tens of 54 
separately, thus, 

54 2 = 50 2 + 2 x 50 x 4 + 4* 

50+ 4 

50 2 X 4 + 2 X 50 x 4 2 + 4 3 
50 3 + 2 X 50 2 x 4 + 50 X 4 2 

54 3 = 50 3 + 3 X 50* x 4 + 3 x 50 X 4 2 + 4 3 = 156924 

Of these combinations, the first is the cube of 50, the second is 3 
times the square of 50 multiplied by 4, the third is 3 times 50 multi- 
plied by the square of 4, and the fourth is the cube of 4. Hence, 

III. The cube of a number composed of tens and units is equal 
to the cube of the tens, plus three times the square of the tens multi- 
plied, by the units, plus three times the tens multiplied, by the square 
of the units, plus the cube of the units. 

By observing the manner in which the cube is formed, we perceive 
that each of the last three parts contains the units as a factor; these 



CUBE root. 385 

parts, considered as one number, may therefore be separated into two 
factors, thus, 

(3 x 50 2 + 3 X 50 X 4 + 4 2 ) X 4. Hence, 

IV. If the cube of the tens be subtracted from the entire cube, 
the remainder will be composed of two factors, one of which will be 
three times the square of the tens plus three times the tens multiplied 
by the units plus the square of the units ; and the other, the units 

1. What is the cube root of 145780726447 ? 

OPERATION. 

145780726447 ( 5263, Ans. 
I II 125 



152 304 



1566 9396 



7500 20780 
780_4__15608_ 

811200 5172726 
820596 4923576 



83002800 249150447 
15783 47349 83050149 249150447 

Analysis. Pointing off the given number into periods of 3 figures 
each, the four periods show that there will be four figures in the root, 
(I). Since the cube of the first figure of the root is contained wholly 
in the first period of the power, (II), we seek the greatest cube in the 
first period, 145, which we find by trial to be 125, and we place its 
root, 5, for the first figure of the required root, and regard it as tens 
of the next inferior order, (654). We now subtract 125, the cube 
of this figure, from the first period, 145, and bringing down the next 
period, obtain 20780 for a remainder. And since the cube of the first 
two figures of the root is contained wholly in the first two periods of 
the power, (Ii), the remainder, 20780, must contain at least the pro- 
duct of the two factors, one of which is three times the square of the 
first figure (tens), plus three times the first figure multiplied by the 
second (units), plus the square of the second ; and the other, the second 
figure (IV) Now if we could divide this remainder by the first of 
these factors, the quotient would be the other factor, or the second 
figure of the root. But as the first factor is composed in part of the 
second figure, which we have not yet found, we can not now obtain the 
complete divisor ; and we therefore write three times the square of 
the first figure, regarded as tens, or 50 2 X3 = 7500, at the left of the 
dividend, for a trial divisor. Dividing the dividend by the trial 
divisor, we obtain 2 for the second, or trial figure of the root. To 
33 z 



386 EVOLUTION. 

complete the divisor, we must add to the trial divisor, as a correction, 
three times the tens of the root already found multiplied by the units, 
plus the square of the units, (IV). But as 50 X 3 X 2 + 2 2 = (50 X 
3 + 2) X 2, we annex the second figure, 2, to three times the first 
figure, 5, and thus obtain 50 X 3 + 2 = 152, the first factor of the 
correction, which we write in the column marked I. Multiplying 
this result by the 2, we have 304, the correction, which we write in 
the column marked II. Adding the correction to the trial divisor, we 
obtain 7804, the complete divisor. Multiplying the complete divisor 
by the trial figure of the root, subtracting the product from the 
dividend, and bringing down the next period, we have 20780 for a 
remainder. 

We have now taken the cube of the first two figures of the root 
considered as tens of the next inferior order, from the first three 
periods of the number ; and since the cube of the first three figures 
of the root is contained wholly in the first three periods of the power, 
(II), the remainder, 20780, must contain at least the product of the 
two factors, one of which is three times the square of the first two 
figures of the root (regarded as tens of the next order) plus three, 
times the first two figures multiplied, by the third, plus the square of the 
third ; and the other, the third figure, (IV). Therefore, to obtain the 
third figure, we must use for a trial divisor three times the square of 
the first two figures, 52, considered as tens. And we observe that the 
significant part of this new trial divisor may be obtained by adding 
the last complete divisor, the last correction, and the square of the 
last figure of the root, thus: 

7804 = (50 2 X 3) + (50 X 3 X 2) + 2 2 

304 = 50 X 3 X 2 + 2 2 

_4== _2 2 

8112 == (50 2 ~ "+ l00~X 2~x +2*) X 3 = 52 2 X 3 
This number is obtained in the operation without re-writing the 
parts, by adding the square of the second root figure mentally, and 
combining units of like order, thus : 4, 4, and 4 are 12, and we write 
the unit figure, 2, in the new trial divisor ; then 1 to carry and 
is 1 ; then 3 and 8 are 11, etc. Annexing two ciphers to the 8112, 
because 52 is regarded as tens of the next order, and dividing by this 
hew trial divisor, 811200, we obtain 6, the third figure in the root. 
To complete the second trial divisor, after the manner of completing 
the first, we should annex the third figure of the root, 6, to three 
times the former figures, 52, for the first factor of the correction. 



CUBE root. 387 

But as we have in column I three times 5 with the 2 annexed, or 152, 
we need only multiply the last figure, 2, by 3, and annex the third 
figure of the root, 6, which gives 1566, the first factor of the correc- 
tion sought, or the second term in column I. Multiplying this number 
by the 6, we obtain 9396, the correction sought ; adding the correction 
to the trial divisor, we have 820596, the complete divisor ; multiplying 
the complete divisor by the 6, subtracting the product from the divi- 
dend, and bringing down the next period, we have 249150947 for a 
new dividend We may now regard the first three figures of the root, 
526, as tens of the next inferior order, and proceed as before till the 
entire root, 5263, is extracted. 

661* From these principles and illustrations we deduce the 
following 

E.ULE. I. Point off the given number into periods of three 
figures each , counting from units place toward the left and right. 

II. Find the greatest cube that does not exceed the left hand 
period, and write its root for the first figure in the required root; 
subtract the cube from the left hand period, and to the remainder 
bring down the next period for a dividend. 

III. At the left of the dividend write three times the square of 
the first figure of the root, and annex two ciphers, for a trial di- 
visor ; divide the dividend by the trial divisor, and write the quo- 
tient for a trial figure in the root. 

IV. Annex the trial figure to three times the former figure, and 
write the result in a column marked J, one line below the trial 
divisor, multiply this term by the trial figure, and write the 
product on the same line in a column marked II ; add this term 
as a correction to the trial divisor, and the result will be the com- 
plete divisor. 

V. Multiply the complete divisor by the trial figure ; subtract 
the product from the dividend, and to the remainder bring down 
the next period for a new dividend. 

VI. Add the square of the last figure of the root, the last term 
in column II, and the complete divisor together, and annex two 
ciphers, for a new trial divisor} with which obtain another trial 
figure in the root. 



388 EVOLUTION. 

VII. Multiply the unit figure of the last term in column 1 by 
3, and annex the trial figure of the root for the next term of 
column I ; multiply this result hy the trial figure of the root for 
the next term of column II; add this term to the trial divisor for 
a complete divisor, with which proceed as before. 

Notes. — 1. If at any time the product be greater than the dividend, diminish 
the trial figure of the root, and correct the erroneous work. 

2. If a cipher occur in the root, annex two more ciphers to the trial divisor, 
and another period to the dividend; then proceed as before with column I, an- 
nexing both cipher and trial figure. 

EXAMPLES FOR PRACTICE. 

1. What is the cube root of 389017 ? Ans. 73. 

2. What is the cube root of 44361864 ? Ans. 354. 

3. What is the cube root of 10469350803 ? Ans. 2187. 

4. What is the cube root of 98867482624 ? Ans. 4624, 

5. What is the cube root of 30.625 ? Ans. 3.12866 +. 

6. What is the cube root of 111* ? Ans. 4.8076 + , 

7. What is the cube root of .000148877 ? Ans, .053. 
Find the values of the following expressions. 

8. V 122615327232? Ans. 4968. 
9- ^ / v/134217W? Ans. 8. 

10. V39304*? Ans. 1156. 

- L1 - ^ 3000 A ^ 4913* .- tt/ **' 8^- 

12 How much does the sum of the cube roots of 50 and 31 
exceed the cube root of their sum? Ans. 2.4986 +. 

CONTRACTED METHOD. 

002. In applying contracted decimal division to the extrac- 
tion of the cube root of numbers, we observe, 

1st. For each new figure in the root, the terms in the operation 
extend to the right 3 places in the column of dividends, 2 places 
in the column of divisors, and 1 place in column I. Hence, 

2d. If at any point in the operation we omit to bring down new 
periods in the dividend, we must shorten each succeeding divisor 
1 place, and each succeeding term in column I, 2 places. 

1. What is the cube root of 189, correct to 8 decimal places ? 



CUBE ROOT. 



389 



II 



OPERATION. 

|5.7 3879355d=, A rts, 

189.000000 
125 



157 


1099 

5139 

1375 

12 


7500 
8599 


64 000 
60193 


1713 


974700 

979839 


3 807000 
2 939517 


1719 


984987 
986362 


867483* 
789090 


17 


98774 
98786 


78393 
69150 






9880 


9243 

8892 




988 


351 
296 




99 


55 
50 




10 


5 
5 



Analysis. We 
proceed by the usual 
method to extract 
the cube root of the 
given number until 
we have obtained 
the three figures, 
5.73 : the corres- 
ponding remainder 
is 867483, and the 
next trial divisor 
with the ciphers 
omitted^is 984987. 
We now omit to 
bring down a period 
of ciphers, thus con- 
tracting the divid- 
end 3 places; and 
we contract the di- 
visor an equal num- 
ber of places by 
omitting to annex the two ciphers, and regarding the right hand 
figure, 7, as a redundant figure. Then dividing, we obtain 8 for the 
next figure of the root. To complete the divisor, we obtain a correc- 
tion, 1375, contracted 2 places by omitting to annex the trial figure 
of the root, 8, to the first factor, 1719, and regarding the right hand 
figure, 9, as redundant in multiplying. Adding the contraction to 
the contracted divisor, we have the complete divisor, 986362, the right 
hand figure being redundant. Multiplying by 8 and subtracting the 
product from the dividend, we have 78393 for a new dividend. Then 
to form the new trial divisor, we disregard the square of the root 
figure, 8, because this square consists of the same orders of units as 
the two rejected places in the divisor; and we simply add the cor- 
rection, 1375, and the complete divisor, 986362, and rejecting 1 figure, 
thus obtain 98774, of which the right hand figure, 4, is redundant. 
Dividing, we obtain 7 for the next root figure. Eejecting 2 places 
from the last term in column I, we have 17 for the next contracted 
term in this column. We then obtain, by the manner shown in the 
former step, the correction 12, the complete divisor, 98786, the prod- 
uct, 69150, and the new dividend, 9243. We then obtain the new trial 
33* 



390 EVOLUTION. 

divisor, 9880; and as column I is terminated by rejecting the two 
places, 17, we continue the contracted divisor as in square root, and 
thus obtain the entire root, 5.73879355 ±, which is correct to the last 
decimal place, and contains as many places as there are places in the 
periods used. Hence the following 

Rule. I. If necessary , annex ciphers to the given number, and 
assume as many figures as there are places required in the root; 
then proceed by the usual method until all the assumed figures have 
been employed. 

II. Form the next trial divisor as usual, but omit to annex the 
two ciphers, and reject one place in forming each subsequent trial 
divisor. 

Ill In completing the contracted divisors, omit at first to annex 
tlie trial figure of the root to the term in column I, and reject 2 
pjaces in forming each succeeding term in this column. 

IV. In multiplying, regard the right hand figure of each con- 
tracted term, in column land in the column of divisors, as redund- 
ant. 

Notes. — 1. After the contraction commences, the square of the last root figure 
is disregarded in forming the new trial divisors. 
2. Employ only full periods in the number. 

EXAMPLES FOR PRACTICE. 

1. Find the cube root of 24, correct to 7 decimal places. 

Arts. 2.8844992 db. • 

2. Find the cube root of 12000.812161, correct to 9 decimal 
places. Ans. 22.894864927 db. 

3. Find the cube root of .171467, correct to 9 decimal places. 

Ans. .555554730 =b. 

4. Find the cube root of 2.555, correct to 5 decimal places. 

Ans. 1.34442±. 

5. Find the cube root of 20.44, correct to 5 decimal places. 

Ans. 2.68884 ±. 

6. Find the value of V\ to 6 places. Ans. .941035 ±. 

7. Find the value of ^.571428 to 9 places. 

Ans. .829827017 ±. 



CUBE ROOT. 391 

8. F.ind the value of Vl.08674325 2 to 7 places. 

Ans. 1.057023 =fc. 

9. Eind the value o.f 1.05 3 to 7 places. 

Ans. 1.0841715 ±. 

ROOTS OF ANY DEGREE. 

603. Any root whatever may be extracted by means of the 
square and cube roots, as will be seen in the two cases which follow. 

CASE I. 

604. When the index of the required root contains 
no other factor than 2 or 3. 

We have seen that if we raise any power of a given number to 
any required power, the result will be that power of the given 
number denoted by the product of the two indices, (64t3, III). 
Conversely, if we extract successively two or more roots of a given 
number, the result must be that root of the given number denoted 
by the product of the indices. 

1. What is the 6th root of 2176782336 ? 

operation. Analysis. The index of the 

6 = 2x3 required root is 6 = 2x3; we 

V 2176782336 = 46657 therefore extract the square root 

V 46656 = 36, Ans. of tne g iven number, and the 

cube root of this result, and ob- 
9 tain 36, which must be the 6th 

^2176782336 = 1296 root required. Or, we first find 

\/l296 = 36, Ans. the cube root of the given num- 

ber, and then the sqxiare root of 
the result, as in the operation. Hence the following 

Rule. Separate the index of the required root into its prime 
factors, and extract successively the roots indicated by the several 
factors obtained; the final result will be the required root. 

examples for practice. 

1. What is the 6th root of 6321363049 ? Ans. 43. 

2. What is the 4th root of 5636405776 ? Ans. 274. 



392 EVOLUTION. 

3. What is the 8th root of 1091123019776? Am. 32. 

4. What is the 6th root of 25632972850442049 ? Am. 542. 

5. What is the 9th root of 1.577635? Am. 1.052567+. 

Note. — Extract the cube root of the cube root by the contracted method, 
carrying the root in each operation to 6 decimal places only. 

6. What is the 12th root of 16.3939 ? Am. 1.2155+ . 

7. What is the 18th root of 104.9617 ? Am. 1.2950+. 

CASE II. 

665. When the index of the required root is prime, 
or contains any other factor than 2 or 3. 

To extract any root of a number is to separate the number into 
as many equal factors as there are units in the index of the re- 
quired root ; and it will be found that if by any means we can 
separate a number into factors nearly equal to each other, the 
average of these factors, or their sum divided the number of fac- 
tors, will be nearly equal to the root indicated by the number of 
factors. 

1. W T hat is the 7th root of 308 ? 

operation. Analysis. We first 

V308 = 2.59+ find ky Case *> tne 6tn 

V3O8 __ 2.04+ root > an< * a l so the 8th 

2.59 + 2.04 = 4^63 root of 308 ; and since 

4.63 -T- 2 = 2.31, assumed root. the 7th root must be 

2.31 6 =151.93 less than the former 

308 -f- 151.93 = 2.0272+ and greater than the 

^L?9 6 . + 7 2 -°9l^ 15 l 8 f 72 ■ x. latter, we take the ave- 
15.8872 -7- 7 = 2.2696, 1st approximation. « xl 
*-£ rage of the two, or one 

2.269 6 = 136.6748 half of their sums 2 31 

308 -r- 136.6748 = 2.253452+ ^ ot ™" ram8 > 2 - sl > 

2.269 X 6 + 2.253452 = 15.871052 and cal1 xt the Mme ^ 

15.871052 -7- 7 = 2.267293, 2d approx. root. We next raise 

the assumed root, 2.31, 
to the 6th power, and divide the given number, 308, by the result, 
and obtain 2.0272+ for a quotient ; we thus separate 308 into 7 fac- 
tors, 6 of which are equal to 2.31, and the other is 2.0272. As these 
7 factors are nearly equal to each other, the average of them all must 
be a near approximation to the 7th root. Multiplying the 2.31 by 6, 
adding the 2.0272 to the product, and dividing this result by 7, we 



cube root. 393 

find the average to be 2.2696, which is the first approximation to the 
required root. "We next divide 308 by the 6th power of 2.2696, and 
obtain 2.253452+ for a quotient; and we thus separate the given 
number into 7 factors, 6 of which are each equal to 2.2696, and the 
other is 2.253452. Finding the average of these factors, as in the 
former steps, we have 2.267293, which is the 7th root of the given 
number, correct to 5 decimal places. Hence the following 

Rule. I. Find by trial some number nearly equal to the re- 
quired root , and call this the assumed root. 

II. Divide the given number by that power of the assumed root 
denoted by the index of the required root less 1 ; to this quotient 
add as many times the assumed root as there are units in the 
index of the required root less 1, and divide the amount by the 
index of the required root. The result will be the first approxi- 
mate root required. 

III. Take the last approximation for the assumed root, with 
which proceed as with the former, and thus continue till the re- 
quired root is obtained to a sufficient degree of exactness. 

Notes. — 1. The involution and division in all cases will be much abridged by 
decimal contraction. 

2. If the index of the required root contains the factors, 2 or 3, we may first 
extract the square or cube root as many times, successively, as these factors are 
found in the index, after which we must extract that root of the result which is 
denoted by the remaining factor of the index. Thus, if the 15th root were re- 
quired, we should first find the cube root, then the 5th root of this result. 

EXAMPLES FOR PRACTICE. 

1. What is the 20th root of 617 ? 

OPERATION. 

20 = 2 x 2 x 5. 
v^617 = 24.839485+. 
v' 24.839485 = 4.983923+. 
v/4.983923 = 1.378838 + . Arts. 

2. What is the 5th root of 120 ? 

3. What is the 7th root of 1.95678 ? 

4. What is the 10th root of 743044 ? 

5. What is the 15th root of 15 ? 
6 What is the 25th root of 100 ? 
7. What is the 5th root of 5 ? 



394 SERIES. 

SERIES. 

666. A Series is a succession of numbers so related to each other, 
that each number in the succession may be formed in the same man- 
ner from one or more preceding numbers. Thus, any number in the 
succession 2, 5, 8, 11, 14, is formed from the preceding number by 
adding 3. Hence, 2, 5, 8, 11, 14, is a series. 

667 The Law of a Series is the constant relation existing be- 
tween two or more terms of the series. Thus, in the series 3, 7, 11, 
15, we observe that each term after the first is greater than the pre- 
ceding term by 4 ; this constant relation between the terms is the law 
of this series. 

The law of a series, and the number of terms on which it depends 
being given, any number of terms of the series can be formed. Thus, 
let 64 be a term of a series whose law is, that each term is four times 
the preceding term. The term following 64 is 64 X 4, the next term 
64 X 4 2 , etc. ; the term preceding 64 is 64 -f- 4. Hence the series, as 
far as formed, is 16, 64, 256, 1024. 

668 A series is either Ascending, or Descending, according as 
each term is greater or less than the preceding term. Thus, 2, 6, 10, 
14, is an ascending series ; 32, 16, 8, 4, is a descending series. 

669. An Extreme is either the first or last term of a series. Thus, 
in the series 4, 7, 10, 13, the first extreme is 4, the last 13. 

670. A Mean is any term between the two extremes. Thus, in 
the series 5, 10, 20, 40, 80, the means are 10, 20, and 40. 

671. An Arithmetical or Equidifferent Progression is a series 
whose law of formation is a common difference. Thus, in the arith- 
metical progression, 3, 7, 11, 15, 19, each term is formed from the 
preceding by adding the common difference, 4. 

672. An arithmetical progression is an ascending or descending 
series, according as each term is formed from the preceding term by 
adding or subtracting the common difference. Thus, the ascending 
series, 7, 10, 13, 16, etc., is an arithmetical progression in which the 
common difference, 3, is constantly added to form each succeeding 
term; and the descending series, 20, 17, 14, 11, 8, 5, -2, is an arith- 
metical progression in which the common difference is constantly 
subtracted, to form each succeeding term. 

673. A Geometrical Progression is a series whose law of forma- 
tion is a common multiplier. Thus, in the geometrical progression, 
3, 6, 12, 24, 48, each term is formed by multiplying the preceding term 
by the common multiplier, 2. 



ARITHMETICAL PROGRESSION. 395 

674. A geometrical progression is an ascending or descending series, 
according as the common multiplier is a whole number or a fraction. 
Thus, the ascending series, 1, 2, 4, 8, 16, etc., is a geometrical pro- 
gression in which the common multiplier is 2 ; and the descending 
series, 32, 16, 8, 4, 2, 1, J, J, etc., is a geometrical progression in 
which the common multiplier is £. 

675. The Ratio in a geometrical progression is the common 
multiplier. 

676. In the solution of problems in Arithmetical or Geometrical 
Progression, five parts or elements are concerned, viz : 

In Arithmetical Progression — 



1. 


The first term ; 


2. 


" last term ; 


3. 


" number of terms ; 


4. 


" common difference; 


5. 


" sum of the series. 



In Geometrical Progression — 
1. The first term ; 


2. 


a 


last term ; 


3. 


a 


number of terms ; 


4. 


a 


ratio ; 


5. 


u 


sum of the series. 



The conditions of a problem in progression may be such as to re- 
quire any one ofMhe five parts from any three of the four remain- 
ing parts ; hence, in either Arithmetical or Geometrical Progres- 
sion, there are 5 X 4 = 20 cases, or classes of problems, and no more, 
requiring each a different solution. 

GENERAL PROBLEMS IN ARITHMETICAL PROGRESSION. 
PROBLEM I. 

677, Given, one of the extremes, the common difference, and 
the number of terms, to find the other extreme. 

Let 2 be the first term of an arithmetical progression, and 3 the com- 
mon difference ; then. 

2 =2 == 2, 1st term. 

2 + 3 =2+ (3 X 1)==. .5, 2d " 

2+3+3 —.2+(3x2)= 8,3d « 

2 + 3 + 3 + 3 = 2 + (3 x 3) == 11, 4th « 

From this illustration we perceive that in an arithmetical progression, 

-when the series is ascending, the second term is equal to the first term plus 

the common difference ; the third term is equal to the first term plus 2 

times the common difference ; the fourth, term is equal to the first term 

£ lus 3 times the common difference ; and so on. In a descending series, 

the second term is equal to the first term minus the common difference ; 

the third term is equal to the first minus 2 times the common difference ; 

and so on. In all cases the difference between the two extremes is equal 



396 



SERIES. 



to the product of the common difference by the number of terms less 1. 
Hence the 

Eule. Multiply the common difference by the number of terms less 
1 ; add the product to the given term if it be the greater extreme, and 
subtract the product from the given term if it be the less extreme. 

EXAMPLES FOR PRACTICE. 

1. The first term of an arithmetical progression is 5, the common dif- 
ference 4, and the number of terms 8 ; what is the last term ? Ans. 33. 

2. If the first term of an ascending series be 2, and the common differ- 
ence 3, what is the 50th term ? 

3. The first term of a descending series is 100, the common difference 
7, and the number of terms 13 ; what is the last term ? 

4. If the first term of an ascending series be f , the common difference 
|, and the number of terms 20, what is the last term ? Ans. 7££ . 

PROBLEM II. 

678. Given, the extremes and number of terms, to find the 
common difference. 

Since the difference of the extremes is always equal to the common 
difference multiplied by the number of terms less 1, (677), we have the 
following 

Rule. Divide the difference of the extremes by the number of terms 
less 1. 

EXAMPLES FOR PRACTICE. 

1. If the extremes of an arithmetical series are 3 and 15, and the num- 
ber of terms 7, what is the common difference ? Ans. 2. 

2. The extremes are 1 and 51, and the number of terms is 75; what is 
the common difference ? 

3. The extremes are .05 and .1, and the number of terms is 8; what is 
the common difference ? Ans. .00625. 

4. If the extremes are and 2 J and the number of terms is 18, what is 
the common difference ? 

PROBLEM III. 

679. Given, the extremes, and common difference to find the 
number of terms. 

Since the difference of the extremes is equal to the common difference 
multiplied by the number of terms less 1, (677), we have the following 

Rule. Divide the difference of the extremes by the common difference, 
and add! to the quotient. 



GEOMETRICAL PROGRESSION. 397 

EXAMPLES FOR PRACTICE. 

1. The extremes of an arithmetical series are 5 and 75, and the common 
difference is 5; what is the number of terms? Ans. 15. 

2. The extremes are J and 20, and the common difference is 6^ ; find the 
number of terms. 

3. The extremes are 2.5 and .25, and the common difference is .125; 
what is the number of terms ? 

4. Insert 5 arithmetical means between 2 and 37. 

PROBLEM IV. 

680. Given, the extremes and number of terms, to find the 
sum of the series. 

Let us take any series, as 2, 5, 8, 11, 14, and writing under it the same 
series in an inverse order, add each term of the inverted series to the term 
above it in the direct series, thus : 

2+ 5+ 8 + 11 + 14 = 40 once the sum, 
I44-H4. 8+ 5+ 2 = 40 « " « 

16 + 16 + 16 + 16 + 16 = 80 twice the sum. 
From this we perceive that 16, the sum of the extremes of the given 
series, multiplied by 5, the number of terms, equals 80, which is twice the 
sum of the series ; and 80 -f- 2 = 40, the sum of the series. Hence 

Rule. Multiply the sum of the extremes by the number of terms, and 
divide the product by 2. 

EXAMPLES FOR PRACTICE. 

1. Find the sum of the series the first term of which is 4, the common 
difference 6, and the last term 40. Ans. 154. 

2. The extremes are and 250, and the number of terms is 1000 ; what 
is the sum of the series ? 

3. A person wishes to discharge a debt in 11 annual payments such that 
the last payment shall be $220, and each payment greater than the pre- 
ceding by $17 ; find the amount of the debt, and the first payment. 

Ans. First payment, $50. 

681. By reversing some one of the four problems now given, or by 
combining two or more of them, all of the sixteen remaining problems 
of Arithmetical Progression may be solved or analyzed. 

GENERAL PROBLEMS IN GEOMETRICAL PROGRESSION. 
PROBLEM I. 

682. Given, one of the extremes, the ratio, and the number 
of terms, to find the other extreme. 
34 



398 SERIES. 

Let 3 be the first term of a geometrical progression, and 2 the common 
difference; then, 

3 =3 — 3, the 1st term. 

3x2 =3 X 2 X = 6, " 2d " 

3x2x2 =3x2 2 =12, "3d " 

3 x 2 X 2 X 2 = 3 x 2 3 = 24, " 4th " 
From this illustration we perceive that in a geometrical progression the 
second term is equal to the first term multiplied by the ratio ; the third term 
is equal to the first term multiplied by the second power of the ratio ; the 
fourth term is equal to the first term multiplied by the third power of the 
ratio ; and so on. The same is true whether the ratio be an integer or a 
fraction. Hence the following 

Kule. I. If the given extreme be the first term, multiply it by that 
power of the ratio indicated by the number of terms less 1 ; the result 
will be the last term, 

II. If the given extreme be the last lerm, divide it by that power of 
the ratio indicated by the number of terms less 1 ; the result will be the 
first term. 

EXAMPLES FOR PRACTICE. 

1. The first term of a geometrical series is 6, the ratio 4, and the num- 
ber of terms 5 ; find the last term. Ans. 6144. 

2. The last term of a geometrical series is 192, the ratio 2, and the 
number of terms 7 • what is the first term ? 

3 If the first term be 6, the ratio J, and the number of terms 8, what 
is the last term ? 

4. The first term is 25, the ratio £, and the number of terms 5 : what is 
the last term I Ans. ^V 

PROBLEM II. 

683. Given, the extremes and number of terms, to find the 
ratio. 

Since the last term is always equal to the first term multiplied by that 
power of the ratio indicated oy the number of terms less 1, (682), we have 
the following 

Rule. Divide the last term by the first, and extract that root of the 
quotient indicated by the number of terms less 1 ; the result will be the 
ratio. 

EXAMPLES FOR PRACTICE. 

1. The extremes are 2 and 512, and the number of terms is 5 ; what is 
the ratio ? Ans. 4. 

2. The extremes are ^ and 45 T 9 ^, and the number of terms is 7 , what 
is the ratio ? 



GEOMETRICAL PROGRESSION. 399 

3. The extremes are 7 and .0112, and the number of terms is 5; what is 
the ratio ? Ans. 5. 

4. Insert 3 geometrical means between 8 and 5000. 

PROBLEM III. 

684. Given, the extremes and ratio, to find the number of 
terms. 

Since the quotient of the last term divided by the first term is equal to 
that power of the ratio indicated by the number of terms less 1, (683), we 
have the following 

Rule. Divide the last term by the first, divide this quotient by the 
ratio, and the quotient thus obtained by the ratio again, and so on in 
successive division, till the final quotient is 1. The number of times 
the ratio is used as a divisor, plus 1, is the number of terms. 

EXAMPLES FOR PRACTICE. 

1. The extremes are 2 and 1458, and the ratio is 3 ; what is the number 
of terms ? Arts. 7. 

2. The first term is .1, the last term 100, and the ratio 10; find the 
number of terms ? 

3. The first term is ^\^, the last term -J, and the ratio 2; what is the 
number of terms ? 

4. The extremes are 196608 and 6, and the ratio is ^; what is the num- 
ber of terms ? Ans. 6. 

PROBLEM IV. 

685. Given, the extremes and ratio, to find the sum of the 
series. 

Let us take the series 5 + 20 + 80 + 320 == 425, multiply each term 
by the ratio 4, and from this result subtract the given series term from 
term, thus : 

20 + 80 + 320 + 1280 = 1700, four times the series, 
5+20+80+329 = 425, once the series, 

1280 — 5 =1275, three times the series, 
Then 1275 +- 3 = 425, once the series. 

Hence the 

Rule. Multiply the greater extreme by the ratio, subtract the less ex- 
treme from the product, and divide the remainder by the ratio less 1. 

Note. — Let every descending series be inverted, and the first term called the 
last; then the ratio will be greater than a unit. If the series be infinite, the least 
term is a cipher. 



400 SERIES. 

EXAMPLES FOR PRACTICE, 

1. The extremes are 3 and 384, and the ratio is 2 ; what is the sum of 
the series ? Ans. 765. 

2. If the extremes are 5 and 1080, and the ratio is 6, what is the sum 
of the series ? 

3. If the first term is 4|, the last term ¥ g 7 , and the ratio J, what is the 
sum of the series ? Ans. 7^. 

4. What is the sum of the infinite series, 8, 4, 2, 1, J, J, etc. ? 

PROBLEM V. 

686. Given, the first term, the ratio, and the number of terms, 
to find the sum of the series. 

If, for example, the first term be 4, the ratio 3, and the number of terms 
6, then by Problem I, we have 

4 X 3 5 = the last term. 
Whence, by Problem IV, we have 

4 X 3« — 4 (3« — 1) X4 ;■ ■ , t . 

— o i — == — o — i = 1456, the sum of the series. 

Hence the following 

Kule. Raise the ratio to a power indicated by the number of terms, 
and subtract \from the result; then multiply this remainder by the first 
term, and divide the product by the ratio less 1. 

EXAMPLES FOR PRACTICE. 

1. The first term is 7, the ratio 3, and the number of terms 4 ; what is 
the sum of the series ? Ans. 280. 

2. The first term is 375, the ratio -£, and the number of terms 4 ; what 
is the sum of the series ? 

3. The first term is 175, the ratio 1.06, and the number of terms 5; 
what is the sum of the series ? 

PROBLEM VI. 

687. Given, the extremes and the sum of the series, to find 
the ratio. 

If we take the geometrical progression 2, 6, 18, 54, 162, in which the 
ratio is 3, then removing the first term and the last term, successively, and 
comparing the results, we have 

6 -f- 18 4- 54 + 162 = sum of the series, minus the first term. 

2 -(- 6 4- 18 + 54 =r sum of the series, minus the last term. 



GEOMETRICAL PROGRESSION. 4Q1. 

Now, since every term in the first line is 3 times the corresponding term 
in the second line, the sum of the terms in the first line must be 3 times 
the sum of the terms in the second line. Hence the 

Kule. Divide the sum of the series minus the first term, by the sum 
of the series minus the last term. 

EXAMPLES FOR PRACTICE. 

1. The extremes are 2 and 686, and the sum of the series is 800 ; what 
is the ratio? Ans. 7. 

2. The extremes are \ and 64, and the sum of the series is 149J ; what 
is the ratio ? 

3. If the sum. of an infinite series be 4J, and the greater extreme 3, 
what is the ratio ? Ans. J. 

688. Every other problem in Geometrical Progression, that admits 
of an arithmetical solution, may be solved either by reversing or com- 
bining some of the problems already given. 

COMPOUND INTEREST BY GEOMETRICAL PROGRESSION. 

689. We have seen (551) that if any sum at compound interest be 
multiplied by the amount of $1 for the given interval, the product 
will be the amount of the given sum or principal at the end of the 
first interval; and that this amount constitutes a new principal for the 
second interval, and so on for a third, fourth, or any other interval. 
Hence a question in compound interest constitutes a geometrical pro- 
gression, whose first term is the principal ; the common multiplier or 
ratio is* one plus the rate per cent, for one interval ; the number of 
terms is equal to the number of years plus one ; and the last term is 
the amount of the given principal for the given time. All the usual 
cases of compound interest, and discount computed at compound 
interest, can therefore be solved by the rules for geometrical progres- 
sion. This will be seen in the following illustration : 

1. Find the amount of $250 for 4 years at 6 % compound interest. 

operation. Analysis. Here 

.40 X 1.0* = $250 X 1.262477 = $316.21925 ^sttr^S £ 

ratio, and 5 the number of terms, to find the last term. Then by 677 we 
find* the last term, which is the amount required. 

EXAMPLES FOR PRACTICE. 

1. What is the amount of $350 in 4 years, at 6 f per annum compound 
interest? Ans. $441.86. 

2. Of what principal is $150 the compound interest in 2 years at 7 4> ? 

34* 2 a 



402 SERIES. * 

3. What sum at 6 $> compound interest will amount to $1000 in 3 years ? 

1 Am. $831.22. 

4. In how many years will $40 amount to $53.24, at 10 ^ compound 
interest ? Am. 5 years. 

5. At what rate per cent, compound interest will any sum double itself 
in 8 years ? Am. 11.04 + <f . 

6. What is the present worth of $322.51, at 5 <f compound interest, due 
24 years hence ? Am. $222.51. 

ANNUITIES. 

690. An Annuity is literally a sum of money which is payable 
annually. The term is however applied to a sum which is payable at 
any equal intervals, as monthly, quarterly, semi-annually, etc. 

Note. — The term, interval, will be used to denote the time between payments. 
Annuities are of three kinds : Certain, Contingent, and Perpetual. 

691. A Certain Annuity is one whose period of continuance is 
definite or fixed. 

692. A Contingent Annuity is one whose time of commencement, 
or ending, or both, is uncertain ; and hence the period of its continu- 
ance uncertain. 

693. A Perpetual Annuity or Perpetuity is one which continues 
forever. 

694. Each of these is subject in reference to its commencement to 
the three following conditions : 

1st. It may be deferred, i. e., it is not to be entered upon until after 
a certain period of time. 

2d. It may be reversionary, i. e., it is not to be entered upon until 
after the death of a certain person, or the occurrence of some certain 
event. 

3d. It may be in possession, i. e., it is to be entered upon at once. 

695. An Annuity in Arrears or Forborne is one on which the 
payments were not made when due. Interest is to be reckoned on 
each payment from its maturity, the same as on any other debt. 

ANNUITIES AT SIMPLE INTEREST. 

696. in reference to an annuity at simple interest, we observe : 

I. The first payment becomes due at the end of the first interval, 
and hence will bear interest until the annuity is settled. 

II. The second payment becomes due at the end of the second in- 
terval, and hence will bear interest for one interval less than the first 
payment. 

III. The third payment will bear interest for one interval less than 
the second, and so on to any number of terms. Hence, 



GEOMETRICAL PROGRESSION". 403 

IV. All the payments being settled at one time, each will be less 
than the preceding, by the interest on the annuity for one interval. 
Therefore they will constitute a descending arithmetical progression, 
whose first term is the annuity plus its interest for as many intervals 
less one as intervene between the commencement and settlement of 
the annuity ; the common difference is the interest on the annuity for 
one interval ; the number of terms is the number of intervals between 
the commencement and settlement of the annuity ; and the last term 
is the annuity itself. 

697. The rules in arithmetical progression will solve all problems 
in annuities at simple interest. 

EXAMPLES FOR PRACTICE. 

1. A man works for a farmer one year and six months, at $20 per 
month, payable monthly; and these wages remain unpaid until the expi- 
ration of the whole term of service. How much is due to the workman, 
allowing simple interest at 6 per cent per annum ? 

operation. Analysis. Here the last 

$20 + $.10 X 17 ±= $21.70, first term. month's wages, $20, is the 

$20 + $21.70 tfrzKon ctlTn last term » the number of 
X 18 — !M'*>.dU, sum. montnS) 18j is the number 

of terms; and the interest 
on 1 month's wages, $.10, is the common difference; and since the first 
month's wages has been on interest 17 months, the progression is a de- 
scending series. Then, by 677 we find the first term, which is the amount 
of the first month's wages for 17 months; and by 680 we find the sum of 
the series, which is the sum of all the wages and interest. 

2. A father deposits annually for the benefit of his son, commencing with 
his tenth birthday, such a sum that on his 21st birthday the first deposit 
at simple interest amounts to* $210, and the sum due his son to $1860. 
How much is the deposit, and at what rate per cent, is it deposited ? 

operation. Analysis. Here the 

$1860X2— $210x12 . nAA , $ 210 > the amount of the 
J2" = $1°°> last term - first deposit, is the first 

210 100 term > 12, the number of 

jj = 10 %, fate. deposits, is the number of 

terms; and $1860, the 
amount of all the deposits and interests, is the sum of the series. By 680 
we find the last term to be $100, which is the annual deposit ; and by 678 
we find the common difference to be $10, which is the annual rate <f * 

3. What is the amount of an annuity of $150 for 5J years, payable 
quarterly, at 1J per cent, per quarter? . Am. $3819.75. 

4. In what time will an annual pension of $500 amount to $3450, at 6 
per cent, simple interest ? Am. 6 years. 



404 SERIES. 

5. Find the rate per cent, at which an annuity of $6000 will amount to 
$59760 in. 8 years, at simple interest? Ans. 7 per cent. 

ANNUITIES AT COMPOUND INTEREST. 

698. An Annuity at compound interest constitutes a geometrical 
progression whose first term is the annuity itself; the common multi- 
plier is one plus the rate per cent, for one interval expressed decimally ; 
the number of terms is the number of intervals for which the annuity 
is taken ; and the last term is the first term multiplied by one plus the 
rate per cent, for one interval expressed decimally raised to a power 
one less than the number of terms. 

699. The Present Value of an Annuity is such a sum as would 
produce, at compound interest, at a given rate, the same amount as 
the sum of all the payments of the annuity at compound interest. 
Hence, to find the present value ; — First find the amount of the an- 
nuity at the given rate and for the given time by (682) ; then find the 
present value of this amount by (685, or 686). 

Notes. — 1. The present value of a reversionary annuity is that principal 
which will amount, at the time the reversion expires, to what will then be the 
present value of the annuity. 

2. The present value of a perpetuity is a sum whose interest equals the annuity. 

700. Questions in Annuities at compound interest can be solved 
by the rules of Geometrical Progression. 

PROMISCUOUS EXAMPLES IN SERIES. 

1. Allowing 6 per cent, compound interest on an annuity of $200 which 
is in arrears 20 years, what is its present amount? Ans. $7357.11. 

2. Find the annuity whose amount for 25 years is $16459.35, allowing 
compound interest at 6 per cent. Ans. $300. 

3. What is the present worth of an annuity of $500 for 7 years, at 6 
per cent, compound interest ? Asns. $2791.18. 

4. What is the present value of a reversionary lease of $100, commen- 
cing 14 years hence, and to continue 20 years, compound interest at 5 per 
cent? Ans, $629,426. 

5. Find the sum of 21 terms of the series, 5, 4|, 4$, etc. 

6. A man travels 13 days ; his last day's journey is 80 miles, and each day 
he travels 5 miles more than the preceding day. How far has he traveled, 
and what was his first day's journey? Ans. He has traveled 650 miles. 

7. Find the 12th term of the series, 30, 15, 7J, etc. Ans. T fa. 

8. The first term of a geometrical progression is 2, the last term 512, 
and common multiplier 4 ; find the sum of the series. Ans. 682. 

9. The distance between two places is 360 miles. In how many days 
can it be traveled, by a man who travels the first day 57 miles, and the last 
45, each day's journey being greater than the preceding by the same num- 
ber of miles ? Ans. 10. 



PROMISCUOUS EXAMPLES. 405 

9. The first term of a geometrical progression is 1, the last term 15625, 
and the number of terms 7 ; find the common ratio. Ans. 5. 

10. An annual pension of $500 is in arrears 10 years. What is the amount 
now due, allowing 6 per cent, compound interest? Ans. $3680.04. 

11. Find the first and last terms of an arithmetical progression whose 
sum is 408, common difference 6, and number of terms 8. 

Ans. First term 30 ; last term 72. 

12. A farmer pays $1196, in 13 quarterly payments, in such a way that 
each is greater than the preceding by $12. What are his first and last 
payments? Ans. $20, and $164. 

13. A man wishes to discharge a debt in such a way that his first pay- 
ment shall be $2, his last $512, and each payment four times the pre- 
ceding payment. How long will it take him to discharge the debt, and 
what is the amount of his indebtedness ? 

14. A man dying, left 5 sons, to whom he gave his property as follows: 
to the youngest he gave $4800, and to each of the others one and a half 
times the next younger son's share. What was the eldest son's fortune, 
and what the amount of property left ? 

Ans. Eldest son's share, $24300; property, $63300. 

15. Find the annuity whose amount for 5 years, at 6 per cent, compound 
interest, is $2818.546. Ans. $2232.552+. 

16. A merchant pays a debt in yearly payments in such a way that each 
payment is 3 times the preceding ; his first payment is $10, and his last 
$7290. What is the amount of the debt, and in how many payments is it 
discharged? Ans. Debt, $11930; 7 payments. 

17. A man traveling along a road stopped at a number of stations, but 
at each station he found it necessary, before proceeding to the next, to 
return to the place whence he started ; the distance from the starting place 
to the fir'st station was 5 miles, and to the last 25 miles ; he traveled in all 
180 miles. How many stations were there on the road, and what was the 
distance from station to station ? Ans. 5 stations ; 4 miles apart. 

18. An annuity of $200 for 12 years is in reversion 6 years. What is 
its present worth, compound interest at 6 <f ? Ans. $2656.039-|-. 

19. A man pays $6 yearly for tobacco, from the age of 16 until he is 
60, when he dies, leaving to his heirs $500. What might he have left 
them, if he had dispensed with this useless habit, and loaned the money at 
the end of each year at 6 ^ compound interest? Ans. $2225.502+. 

20. What is the present worth of a reversionary perpetuity of $100, 
commencing 30 years hence, allowing 5 per cent, compound interest ? 

21. A merchant purchased 8 pieces of cloth, for which he paid $136; 
the difference in the length of any two pieces was 2 yd., and the difference 
in the price $4. He paid $31 for the longest piece, and $1 a yard for the 
shortest. Find the whole number of yards, and the price per yard of each 
piece. 

22. A farmer has 600 bushels of different kinds of grain, mixed in 
such a way that the number of bushels of the several kinds constitute a 
geometrical progression, whose common multiplier is 2 ; the greatest num- 
ber of bushels of one kind is 320. Find the number of kinds of grain 
in the mixture, and the number of bushels of each kind. Ans. 4 kinds. 



406 



MENSURATION. 



MENSURATION OF LINES AND SUPERFICES. 

701. An Angle is the opening between two lines 
that meet each other. 

702. A Right Angle is an angle formed by two 
lines perpendicular to each other. Thus, B A C is a A 
right angle. 

703. If an angle is less than a 
right angle, it is acute; if greater 
than a right angle, it is obtuse. 
Thus, the angle on the right of the 
line C B is acute, and the angle on 
the left of C B is obtuse. 



c 




704. Parallel Lines are lines hav- A 

ing the same direction, as A and B. B 

705. A Triangle is a figure having three sides 
and three angles, as ABC. 

706. A Eight-Angled Triangle is a triangle 
having one right angle, as at C. 

707. The Hypotenuse is the side opposite the 
right angle, as A B. 

708. The Base of a triangle is the side on 
which it is supposed to stand, as A C. 

709. The Altitude of a triangle is the per- 
pendicular distance from the base or the base produced to the angle 
opposite, as C B. 

Note. — The altitude of a right-angled triangle is the side called the perpen- 
dicular. 

710. A Square is a figure having four equal sides and four right 
angles. c 

711. A Rectangle is a figure having four 
right angles, and its opposite sides equal. 

712. A Diagonal is a line drawn through 
a figure, joining two opposite angles, as A C. 

713. A Rhombus is a figure having four equal 
sides and four oblique angles. 

714. A Rhomboid is a figure having its opposite 
sides equal and parallel, and its angles oblique. 

Note. — The square, rectangle, rhombus, and rhomboid, 
having their opposite sides parallel, are called by the general 
name, parallelogram. 




715. A Trapezoid is a figure having four sides, 
two of which are parallel. 



LINES AND SURFACES. 



407 




716. A Circle is a figure bounded by one 
uniform curved line. 

717. The Circumference of a circle is the 
curved line bounding it. 

718. The Diameter of a circle is a straight 
line passing through the center, and termina- 
ting in the circumference. 

PROBLEM I. 

719. To find either side of a right-angled triangle, the other 
two sides being given. 

Let us take any right-angled triangle, 
as A B C, and form the square, A E D C, 
on the hypotenuse. Now take a portion, 
A B C, of this square, and move it as on 
a hinge at A, until the points B and C 
are brought to the positions of H and 
E, respectively. Take also another por- 
tion, DFC, and move it as on a hinge 
at D, until the points F and C are 
brought to the positions of G and E, 
respectively. Then the figure formed 
by the parts thus moved and the re- 
maining part will be composed of two new squares, one on A B, the base 
of the triangle, and one on D F, which is equal to the perpendicular of 
the triangle. Hence, 

The square of the hypotenuse of a right-angled triangle is equal to 
the sum of the squares of the other two sides. 

From this property we derive the following 

Rule. I. To find the hypotenuse; — Add the squares of the two 
sides, and extract the square root of the sum. 

II. To find either of the shorter sides ; — Subtract the square of the 
given side from the square of the hypotenuse, and extract the square 
root of the remainder. 




EXAMPLES FOR PRACTICE. 

1. The top of a tower standing 22 feet from the shore of a river, is 75 
feet above the water, and 256 feet in a straight line from the opposite 
shore; required the width of the river. Ans. 223.76 ft. 

2. Two ships set sail from the same port, and one sails due east 50 
leagues, the other due north 84 leagues ; how far are they apart? 

3. A ladder 50 ft. long will reach a window 30 ft. from the ground on 



408 MENSURATION. 

one side of the street, and without moving the foot will reach a window 
40 ft. high on the other side ; what is the breadth of the street? 

4. What is the distance through a cubical block, measured from one 
corner to the opposite diagonal corner, the side of the cube being 6 feet ? 

5. There is a park containing an area of 10 A. 2 R. 20 P., and the 
breadth is f of its length. If two men start from one corner and travel 
at the rate of 3 miles per hour, one going by the walk around the park, 
and the other taking the diagonal path through the park, how much 
sooner will the latter reach the opposite corner than the former. 

Ans. 1 min. 52J sec. 

720. The Area of a figure is its superficial contents, or the sur- 
face included within any given lines, without regard to thickness. 

In taking the measure of any line, or surface, we are always gov- 
erned by some denomination, a unit of which is called the Unit of 
Measure, (279). 

PROBLEM II. 

721. To find the area of a square or rectangle. 

Rule. Multiply the length by the breadth. 
Note. — For analysis of the principles of this rule see 

EXAMPLES FOR PRACTICE. 

1. How many square yards in the floor, ceiling, and walls of a 
room 24 feet long, 15 feet wide, and 8 J feet high ? Ans. 153f . 

2. The boundary lines of my farm taken in order are as follows : 
The first runs north 38 ch. 20 1. ; the second, east 25 ch. 14 1. ; the 
third, south 12 ch. 8 1. ; the fourth, west 8 ch. 30 1. ; the fifth, south 
26 ch. 12 1. ; the sixth, west 16 ch. 84 1. to the place of beginning. 
Required the area. Ans. 74 A. -66.4+ P. 

3. If a piece of land 20 rods long contain 240 square rods, what is 
its width ? Note. — Reverse the rule. 

4. A surveyor wishes to lay out a rectangular lot of land which 
shall be 120 rods in length, and contain 70 acres ; what must be its 
width? Ans. 93 i rods. 

5. A piece of land 8 chains wide contains 40 acres ; what is its 
length in chains? Ans. 50 chains. 

PROBLEM III. 

722. To find the area of a rhombus or a rhomboid. 

Rule. Multiply the length by the shortest or perpendicular distance 
between two opposite sides. 

EXAMPLES FOR PRACTICE. 

1. The side of a plat of ground in the form of a rhombus is 36 feet, 
and the perpendicular distance between either two of the sides is 28 
ft. 9 in. ; what is the area of the plat? Ans. 3 sq. rd. 218 J sq. ft. 






LINES AND SURFACES. 409 

2. The longer sides of a rhomboid measure 72 rods, and the shorter 
sides 20 rods ; and a perpendicular from the obtuse angle at the ex- 
tremity of one of the longer sides meets the opposite side 12 rods from 
the acute angle at its extremity. Required the area of the figure. 

Ans. 7 A. 22 P. 

3. There is a piece of land in the form of a rhomboid, whose longer 
sides run north and south, and are 30 rods apart ; and whose shorter 
sides are half the length of the longer ; if a piece of land in rectangu- 
lar form was enclosed by lines of the same length as these, respect- 
ively, it would contain 14 A. 72 P. What is the area of the rhom- 
boid? Ans. 12 A. 120 P. 

PROBLEM IV. 

723. To find the area of a trapezoid. 

Rule . Multiply one-half the sum of the parallel sides by the perpen- 
dicular distance between them. 

EXAMPLES FOR PRACTICE. 

1. My farm has four sides, two of which are parallel, and at a dis- 
tance from each other of 30 ch .25 1. ; the lengths of the parallel sides 
are 72 ch. 40 L, and 84 ch. 36 1., respectively. How many acres in my 
farm? Ans. 237 A. 43.36 P. 

2. What is the area of a board 12 feet long, 16 inches wide at each 
end, and 9 in the middle ? 

3. The boundary lines of a meadow are as follows : the first runs 
north 36 rods, the second in a northeasterly direction 30 rods, the 
third south 54 rods, and the fourth west 24 rods, to the place of be- 
ginning. Required the area of the meadow. Ans. 6 A. 120 P. 

PROBLEM V. 

724. To find the area of a triangle. 

Rule. Multiply one half the base by the altitude, or one-half the 
altitude by the base ; or, 

Multiply the base by the altitude, and divide the product by two. **k^ 

EXAMPLES FOR PRACTICE. 

1. How many square yards in a triangle whose base is 126 feet, 
and perpendicular 24 feet ? Ans. 168 sq. yd. 

2. The gable ends of a barn are each 34 ft. wide, and the perpen- 
dicular hight of the ridge above the eaves is 8 ft. ; how many feet of 
boards will be required to board up both gables ? Ans. 306 ft. 

Note 1. — It will readily be seen that the gable may be divided into two right- 
angled triangles. 

3. The area of a right-angled triangle is 48 feet, and the base is 12 
feet ; what is the perpendicular ? Ans. 8 ft. 

Note 2. — This example is the reverse of the preceding ones. 
35 ' 



410 MENSURATION. 

4. The area of the gable of a certain building is 108 feet, and the 
perpendicular hight of the ridge of the roof above the eaves is 9 feet ; 
what is the width of the building ? 

5. One side of a triangular field is 18 chains in length, and 
the perpendicular distance between this side and the opposite 
angle is 15 chains ; what is the area of the field ? 

Ans. 3 E. 15 P. 

6. "What are the contents of a triangular board, each of whose sides 
measures 18 inches? Ans. 280.44 + sq. in. 

PROBLEM VI. 

725. To find the diameter or the circumference of a circle. 

It is proved in Geometry that in every circle the ratio between the 
diameter and the circumference is 3.14159+ . Hence 

Rule. I. To find the diameter; — Multiply the circumference by 
.31831. 
II. To find the circumference ; — Multiply the diameter by 3.14159. 

EXAMPLES FOR PRACTICE. 

1. What is the circumference of a circle 8 feet in diameter? 

Ans. 25 ft. 1.58+ in. 

2. If the circumference be 49.52 rods, what is the diameter? 

Ans. 15.762 rods. 

3. What is the length of an arc of 18° in a circle whose radius is 4 
ft. 8J in. ? ^ Ans. 1 ft. 5.75 in. 

4. Within a circular garden 66 chains in circumference, is a circu- 
lar pond 66 rods in circumference ; what is the width of the ring of 
land surrounding the pond ? Ans. 31.51+ rods. 

5. The circumference of a cart wheel is 16 ft. 6 in. ; what is the 
diameter? Ans. 5 ft. 3 in. 

PROBLEM VII. 

726. To find the area of a circle. 

Rule. I. When both diameter and circumference are given; — 
Multiply the diameter by the circumference, and divide the product 
by 4. 

II. When the diameter is given ; — Multiply the square of the diam- 
eter by .7854. 

III. When the circumference is given ; — Multiply the square of the 
circumference by .07958. 

Note. — It is demonstrated in Geometry, that the areas of two circles are to 
each other ns the squares of their radii, diameters, or circumferences. Hence, 
when the ratio of any one of these parts in any circle to the corresponding part 
of a known circle, is given, the problem may be solved by proportion. 



SOLIDS. 



411 



EXAMPLES FOR PRACTICE. 

1. The diameter of a circle is 226, and the circumference 710 ; what 
is the area? Ans. 39115. 

2. What is the area of a circular saw 25 inches in diameter ? 

Ans. 3 sq. ft. 58^ sq. in. 

3. The circumference of one end of a log is 6 ft. 10 in. ; what is the 
area? 

4. A circular plat contains 6.44598 square chains of land ; how 
many rods in the circumference ? Ans. 36. 

5. What is the area of a quadrant whose radius is 4 rods ? 

Ans. 12.5664 sq. rd.. 

6. If a cistern 6 feet in diameter holds 80 barrels of water, how 
much water will be contained in a cistern of the same depth 18 feet 
in diameter ? 

7. What is the length of the side of a square which contains the 
same area as a circle 7 feet in diameter? Ans. 6.235+ feet. 

8. What is the length of the side of a square which can just be 
inclosed within a circle 42 inches in diameter ? Ans. 29.7 — in. 

9. If it costs $75 to inclose a circular fish pond containing 3 A. 86 
P., how much will it cost to inclose another containing 17 A. 110 P.? 

Ans. $167.70. 



MENSURATION OF SOLIDS. 

727. A Solid or Body is a magnitude which has length, breadth, 
and thickness. 

728. In estimating the solid contents of a body, we are always 
governed by some denomination, a unit of which is called the Unit of 
Measure, (287). 

729. A Prism is a solid whose bases or 
ends are any similar, equal, and parallel plane 
figures, and whose sides are parallelograms. 

730. A Parallelepiped is a solid bounded 
by six parallelograms, the opposite ones of 
which are parallel and equal to each other. Or, 
it is a prism whose base is a parallelogram. 

731. A Cube is a solid bounded by six 
equal squares. The cube is sometimes called a 
Right Prism. 

732. A Cylinder is a body whose bases or 
ends are equal and parallel circles, and whose 
side is a uniform curved surface. 

733. The Altitude of a prism, cube, paral- 
lelopiped, or cylinder, is the perpendicular dis- 
tance between two bases ; it is the length of the 
body. 




412 MENSURATION. 

734. A Pyramid is a solid whose base is any plane 
figure, and whose sides are triangles terminating in a 
point at the top. 

735, A Cone is a solid whose base is a circle, and 
whose side is a curved surface terminating in a point at 
the top. 



736. The Vertex of a pyramid or cone is the point 
in which the sides terminate. 




737. The Altitude of a pyramid or cone is the per- 
pendicular distance from the vertex to the base. 

Note. — The slant height of a cone is the distance from the ver- 
tex to the circumference of the base ; and the slant height of a 
pyramid is the distance from the vertex to the middle of the side 
of the base. 

738. A Sphere or Globe is a solid bounded 
by a single curved surface, which in every part 
is equally distant from a point within called its 
center. 

739. The Diameter of a sphere is a straight 
line passing through its center, and terminating 
at its surface. 

740. A Hemisphere is one half of a globe or sphere. 

PROBLEM I. 

741. To find the cubic contents of a prism, cube, parallelopiped, 
or cylinder. 

To estimate the solid contents of a prism, parallelopiped, cube, 
or cylinder, suppose one end or base to be divided into square or 
superficial units ; then a cubic or solid unit, of the same denomina- 
tion, within the body, may rest upon each one of these square units, 
and there will be as many such tiers of solid units within the body as 
there are units of length between the two ends, or in the altitude of 
the body. Hence the 

Rule. Multiply the area of the base by the altitude. 

Note 1. — It is proved in Geometry that all similar solids are to each other as 
the cubes of their like dimensions. Hence, any dimension may be found by 
proportion, when its ratio to the corresponding dimension of a known similar 
solid is given. 



solids. 4^3 

examples for practice. 

1. The side of a cubic block measures 16 feet ; how many cubic feet 
does it contain ? Arts, 4096. 

2. What are the solid contents of a parallelopiped 15 feet long, 3 
feet wide, and 11 inches thick ? 

3. The end of a prism 24 feet long is a triangle, each side of which 
is 1 foot long ; what are the cubic contents of the prism ? 

Ans. 10.7+ cu. ft. 

4. A cylindrical cistern is 9 feet in diameter, and 9 J feet deep ; 
how many hogsheads of water will it hold ? 

Arts. 64 hhd. 56.36+ gal. 

5. A log 20 feet long, and 10 inches in diameter, was hewn square; 
how many cubic feet were cut away ? 

Ans. 3 cu. ft. 166.5+ cu. in. 

6. How many gallons of water will be discharged in 1 h. 20 min., 
by a pipe 2 inches in diameter, the velocity of the stream being 12 
inches per second? Ans. 783.36 gal. 

Notes. — 2. The mean or average diameter of a barrel or cask may be found, 
by adding to the head diameter § , or, if the staves be but little curving, y 6 g of 
the difference between the head and bung diameters. The cask will then be re- 
duced to a cylinder, and its contents may be found by the above rule. 

3. The process of estimating the capacity of barrels or casks is called Gauging. 

7. The head diameter of a cask is 23 inches, the bung diameter 29 
inches, and the length 36 inches ; how many standard gallons does it 
contain? Ans. 89.66+. 

8. What is the capacity of a cask, of which the head diameter is 
25 inches, the bung diameter 28 inches, and the length 35 inches ? 

PROBLEM II. 

742. To find the cubic contents of a pyramid or cone. 

It is proved in Geometry that the solid contents of a pyramid or 
a cone are equal to J of the solid contents of a prism or cylinder 
respectively, of the same base and altitude. Hence, 

Rule. Multiply the area of the base by J of the altitude. 

EXAMPLES FOR PRACTICE. 

1. A pyramid is 17 feet square at the base, and 36 feet high ; what 
are its solid contents ? Ans. 3468 cu. ft. 

2. What is the solidity of a cone of which the altitude is 5 feet, 
and the circumference of the base 5}^ ft. ? Ans. 4.64295+ cu. ft. 

3. Out of a cone 18 feet high, and 7 feet in diameter at the base, a 
pyramid was cut having its base square ; required the solidity of the 
pyramid. Ans. 147 cu. ft. 

4. What is the solidity of a pyramid 30 feet square at the base, the 
slant height being 25 ft. ? Ans. 6000 cu. ft. 

5. The slant height of a cone is 18 inches, and the circumference 
of the base 30 inches ; what are the cubic contents ? 

35* 



414 MENSURATION. 

PROBLEM III. 

743. To find the surface or the solid contents of a sphere. 

It is demonstrated in Geometry that the surface of a sphere is 4 
times as great as the area of a circle of the same diameter ; and 
that the solid contents of a sphere are £ of the product of the surface 
and diameter. Hence, 

Eule. I. To find the surface ; — Multiply the square of the diameter 
by 3.1416. 

II. To find the solid contents ; — Multiply the cube of the diameter by 
.5236. 

EXAMPLES FOR PRACTICE. 

1. How many square feet on the surface of a sphere 4 ft. in diame- 
ter? Ans. 50.2656. 

2. What is the solidity of a wicket ball 8 J inches in diameter ? 

Ans. 321.55+ cu. in. 

3. What are the solid contents of a hemisphere 12 inches in dia- 
meter? Ans. 358.11+ cu. in. 

4. What is the length of one side of a cube that can just be in- 
closed within a hollow square 12 ft. in diameter ? Ans. 8.48+ ft. 

5. The solidity of a sphere is 65.45 cu. in. ; what is its diameter? 

6. If a ball 6 in. in diameter weigh 32 lb., what will be the weight 
of one 3 in. in diameter of the same metal ? Ans. 4 lb. 

7. If a globe of silver 3 in. in diameter be worth $75, what is the 
value of one 6 J in. in diameter ? 

PROBLEM IV. 

744. To find the cubic contents of any irregular solid. 

It is a truth of Philosophy that matter is impenetrable, or that two 
bodies cannot occupy the same space at the same time. Hence, 

Rule. Fill any vessel with water, and in this immerse the body 
whose contents are required; the quantity of water displaced will be 
equal to the contents of the body immersed. 

EXAMPLES FOR PRACTICE. 

1. A piece of lead was immersed in a gallon measure of water, by 
which one half the water was displaced ; what was the lead worth, at 
15 cents a pound? Ans. $5,775. 

2. A blacksmith's anvil was put into a tub the capacity of which 
was 8} wine gal., and the tub was afterward filled with 6 gal. 3 qt. 
1 pt. of water ; what was the solidity of the anvil? Ans. 317| cu. in. 

3. A chain was put into a cubical box whose inside measure was 8 
inches, and the box was afterward filled with 3f quarts of water ; what 
were the cubic contents of the chain? Ans. 395 T \ cu. in. 



MISCELLANEOUS EXAMPLES. 415 

MISCELLANEOUS EXAMPLES. 

1. How many thousands of shingles will cover both sides of a roof 
36 ft. long, and whose rafters are 18 ft. in length ? 

2. From f of f of J of 70 miles, subtract .73 of 1 mi. 3 fur. 

3. What number is that from which if 7 J be subtracted, f of the 
remainder is 91 J ? Arts. 144|. 

4. What part of 4 is £ of 6 ? Ans. f . 

5. It is required to mix together brandy at $.80 a gallon, wine at 
$.70, cider at $.10, and water, in such proportions that the mixture 
may be worth $.50 a gallon ; what quantity of each must be used ? 

Ans. 9 gal. each of brandy and wine ; 5 gal. each of cider and water. 

6. What number increased by J, J, and i of itself equals 125 ? 

7. What is the hour, when the time past noon is equal to f of the 
time to midnight ? Ans. 4 h. 48 min. p. m. 

8. A grocer mixed 12 cwt. of sugar @ $10, with 3 cwt. @ $8f, 
and 8 cwt. @ $7 i ; how much was 1 cwt. of the mixture worth ? 

9. If $240 gain $5.76 in 4 mo. 26 da., what is the rate <f ? Ans. 6. 

10. If 24 men, in 189 da., working 14 h. a day, dig a trench 33f yd. 
long, 3 i yd. deep, and 5f yd. wide ; how many hours* a day must 217 
men work to dig a trench 23 J yd. long, 2 J yd. deep, and 3f yd. wide, 
in 5 J days ? Ans. 16 h. 

11. What is the difference between the interest and the discount 
of $450 at 5 per cent., for 6 yr. 10 mo. ? 

12. A younger brother received $6300, which was J as much as his 
elder brother received ; how much did both receive ? Ans. $14400. 

13. Keduce .7, .88, .727, .91325 to their equivalent common fractions. 

14. A person by selling a lot of goods for $428, loses 10 4, ; how 
much should the goods have sold for, to gain 12J fo ? 

15. For what sum must a note be drawn at 4 mo., that the proceeds 
of it, when discounted at bank at 7 per cent., shall be $875.50? 

16. Three persons engaged in trade with a joint capital of $2128 ; 
A's capital was in trade 5 mo., B's 8 mo., and C's 12mo. ; A's share 
of the gain was $228, B's $226.40, and C's $330. What was the capi- 
tal of each ? Ans. A's, $912 ; B's, $666 ; C's, $550. 

17. Henry Truman purchased corn of John Bates, on 2 months' 
credit, as follows: Aug. 27, 300 bu. @ $.35 ; Aug. 31, 150 bu. @ 
$.40; Sept. 7, 500 bu. @ $.38; Sept. 12, 200 bu. @ $.42; Sept. 25, 
250 bu. @ $.40. When was the a|c due per average ? Ans. Sept. 8. 

18. A B and C can do a job of work in 12 da., C can do it in 24 da., 
and A in 34 da. ; in what time can B do it alone ? Ans. 81| da. 

19. If a man travel 7 mi. the first day, and 51 mi. the last, increas- 
ing his journey 4 mi. each day, how many days will he travel, and 
how far ? Ans. 12 da., and 348 mi. 

20. What is the difference between the true and bank discount of 
$2500, payable in 90 days at 7 per cent. ? 

21. Which is the more advantageous, to buy flour at $5 a bbl. on 6 
mo., or $4.87J cash, money being worth 7 <f ? Ans. At $5 on 6 mo. 

22. Sold \ of a lot of lumber for what f of it cost ; what oL was 
gained on the part sold? 



416 MISCELLANEOUS EXAMPLES. 

23. If $500 gain $50 in 1 yr., in what time will $960 gain $60 ? 

24. Eeceived an invoice of crockery, 12 per cent, of which was 
broken ; at what per cent, above cost must the remainder be sold, to 
clear 25 per cent, on the invoice ? Ans. 42^0. 

25. The sum of two numbers is 365, and their difference is .0*875 ; 
what are the numbers ? 

26. If the interest of $44562} be $128.99 for 7 yr., what will be the 
interest of $650 for 3 yr. 10 mo. 15 da.? Ans. $104.15. 

27. Eeceived from Savannah 150 bales of cotton, each weighing 
540 lb., and invoiced at 7d. a pound Georgia currency. Sold it at an 
advance of 26 fo, commission 1} fo, and remitted the proceeds by 
draft. What was the face of the draft, exchange being } <f discount ? 

28. A man in Chicago has 5000 francs due him on account in Paris. 
He can draw on Paris for this amount, and negotiate the bill at 19 1 
cents per franc ; or he can advise his correspondent at Paris to remit 
a draft on the United States, purchased with the sum due him, ex- 
change on U. S. being at the rate of 5 francs 20 centimes per $1. 
What sum will the man receive by each method ? 

Ans. By draft on Paris, $970 ; by remittance from Paris, $961.53. 

29. What sum must be invested in stocks bearing 6^ %, at 105 £, 
to produce an income of $1000 ? 

30. A person exchanges $250 shares of 6 per cent, stock, at 70 % , 
for stock bearing 8 per cent., at 120 fo ; what is the difference in his 
income ? 

31. If | of A ? s money equals § of B's, and § of B ? s equals § of C's, 
and the interest of all their money for 4 yr. 8 mo. at 6 fo is $15190, 
how much money has each ? 

32. A boy 14 years old was left an annuity of $250, which was de- 

Eosited in a savings bank at 6 % interest, payable semi-annually; 
ow much will he be worth when of age ? 

33. If a boy buys peaches at the rate of 5 for 2 cents, and sells 
them at the rate of 4 for 3 cents, how many must he buy and sell to 
make a profit of $4.20? 

34. What % in advance of the cost must a merchant mark his 
goods, so that, after allowing 5 fy of his sales for bad debts, an ave- 
rage credit of 6 months, and 7 % *of the cost of the goods for his ex- 
penses, he may make a clear gaifi>of 12^ % on the first cost of the 
goods, money being worth 6 % ? 

35. Four men contracted to do a certain job of work for $8600 ; the 
first employed 28 laborers 20 da., 10 h. a day; the second, 25 labor- 
ers 15 da., 12 h. a day; the third, 18 laborers 25 da., 11 h. a day; 
and the fourth, 15 laborers 24 da., 8 h. a day. How much should each 
contractor receive ? 

36. If I exchange 75 railroad bonds of $500 each, at 36 % below 
par, for bank stock at 5 fo premium, how many shares of $100 each 
will I receive? 

37. A trader has bought merchandise as follows: July 3, $35.26 ; 
July 4, $48.65, on 30 da. ; Aug. 17, $6.48 ; Sept. 12, $50. What is 
due on the account Oct. 12, interest at 9 °/c ? Ans. $192.0-1. 

38. A farmer sold 34 bu. of corn, and 26 bu. of barley for $63.10, 



MISCELLANEOUS EXAMPLES. 417 

receiving 35 cents a bushel more for the barley than for the corn ; 
what was the price of each per bushel? 

39. A speculator purchased a quantity of flour, Sept. 1 ; Oct. 1 its 
value had increased 25 % ; Nov. 1 its value was 30 °jo more than Oct. 
1 ; Dec. 1 he sold it for 15 fo less than its value Nov. 1, receiving in 
payment a 6 months' note, which he got discounted at a bank, re- 
ceiving $12950 on it. Allowing money to be worth 7 fo 9 how much was 
his profit on the flour ? 

40. A flour merchant bought 120 bbl. of flour for $660, paying 
$5.75 for first quality and $5 for second quality ; how many barrels 
were first quality ? Ans. 80. 

41. Two mechanics work together ; for 15 days' work of the first 
and 8 days' work of the second they receive $61, and for 6 days' 
work of the first and 10 days' work of the second they receive 
$38 ; how much does each man earn ? Ans. 1st, $63 ; 2d, $36. 

42. The duty, at 15 fo, on Rio coffee, in bags weighing 180 lbs. 
gross, and invoiced at $.12} per pound, was $961.87}, tare having been 
allowed at 5 fo ; how many bags were imported ? Ans. 300. 

43. A dairyman took some butter to market, for which he received 
$49, receiving as many cents a pound as there were pounds ; how 
many pounds were there ? 

44. A mechanic received $2 a day for his labor, and paid $4 a week 
for his board ; at the expiration of 10 weeks he had saved $72 ; how 
many days did he work, and how many was he idle ? 

45. To what would $250, deposited in a savings bank, amount in 
10 yr., interest being paid semi-annually at 6 fo per annum ? 

46. How much water must be put with 100 gal. of wine that cost 
$120, so that 1 gal. of the mixture may be sold for $1, without loss ? 

47. If a pipe 3 in. in diameter will discharge a certain quantity of 
water in 2 h., in what time will 3 two-inch pipes discharge 3 times 
the quantity ? 

48. Wm. Jones & Co. become insolvent and owe $8100. Their 
assets amount to $4981.50. "What per cent, of their indebtedness can 
they pay, allowing the assignees 2} fo on the amount distributed 
for their services ? sJl Ans. 60 per cent. 

49. Shipped a car load of fat catkin to Boston, and offered them for 
sale at 25 per cent, advance on th^rcost, but the market being dull I 
sold for 14 per cent, less than my asking price, and gained thereby 
$170. How much did the cattle cost ; for how much did they sell ; 
and what was my asking price ? 

50. What must be the dimensions of a cubical cistern to hold 2000 
gallons ? 

51. A man died leaving $5000 to be divided between his three sons, 
aged 13, 15, and 16 yr. 6 mo., respectively, in such a proportion that 
the share of^ each being put at interest at 6 % , should amount to the 
same sum when they should arrive at the age of 21. How much was 
each one's share ? 

52. A vessel having sailed due south and due east on alternate days, 
was found, after a certain time, to be 118.793 miles south-east of the 
place of starting ; what distance had she sailed Hj Ans. 168 miles. 

53. Imported 4 pipes of Madeira wine, at $2.15 a gallon, and paid 



418 MISCELLANEOUS EXAMPLES. 

$57.60 freight, and a duty of 24 per cent. I sold the whole for $1980 ; 
what was my gain % ? 

54. If 34J bu. of corn are equal in value to 17 bu. wheat, 9 bu. 
wheat to 59J bu. oats, 6 bu. oats to 42 lb. flour, and 56 lb. flour to 
$1.96, how many bushels of corn will purchase 5 bbl. of flour ? 

55. If stock bought at 8 % discount will pay 7 % on the invest- 
ment, at what rate should it be bought to pay 10 % ? 

56. A merchant in New York gave $2000 for a bill of exchange of 
£400 to remit to Liverpool ; what was the rate fo in favor of England ? 

57. A, B, and C start from the same point, to travel around a lake 
84 miles in circumference. A travels 7 miles, and B 21 miles a day 
in the same direction, and C 14 miles in an opposite direction. In 
how many days will they all meet ? Am. 12. 

58. The exact solar year is greater than 365 days by i§||§ of a 
day ; find approximately how often leap year should come, or one day 
be added to the common year, in order to keep the calendar right. 

Arts. Once in every 4 yr. ; 7 times in every 28 yr. ; 8 times in every 
33 yr. ; 31 times in every 128 yr. ; or 163 times in every 673 yr.» 

59. A gentleman purchases a farm for $10000, which he sells after 
a certain number of years for $14071, making on the investment 5 <fo 
compound interest. He now invests his money in a, perpetuity, which 
is in reversion 11 years from the date of purchasing the farm. Al- 
lowing 6 </o compound interest for the use of money, find the annuity 
and the length of time he owns the farm. 

60. What will I gain fo by purchasing goods on 6 mo., and selling 
them immediately for cash at cost, money being worth 7 fo ? 

61. "What sum must a man save annually, commencing at 21 yea,rs 
of age, to be worth $30000 when he is 50 years old, his savings being 
invested at 5 fo compound interest ? 

62. Three sons are to share $10000 in the ratio of 3, 4, and 5, but 
the youngest dying it is required to divide the whole sum equitably 
between the other two. What are the shares of the other two? 

63. If 50 bbl. of flour in Chicago are worth 125 yd. of cloth in New 
York, and 80 yd. of cloth in New York are worth 6 bales of cotton in 
Charleston, and 13 bales of cotton in Charleston are worth 3J hhd. 
of sugar in New Orleans, how many hhd. of sugar in New Orleans 
are worth 1500 bbl. of flour in Chicago ? 

64. Seven men all start together to travel the same way around, an 
island 120 miles in circumference, and continue to travel until they 
all come together again. They travel 5, 6 J, 7£, 8£, 9 J, 10 J and 11} 
miles a day respectively. In how many days will they all be together 
again ? 

65. There are two clocks which keep perfect time when their pen- 
dulums beat seconds. The first loses 20 seconds a day, and the second 
gains 15 seconds a day. If the two pendulums beat together when 
both dials indicate precisely 12 o'clock, what time will each clock 
show when the pendulums next beat in concert ? 

Am. The first shows 41 min. 8 sec. past 12; and the second 41 
min. 9 sec. past 12. 

_ 66. If a body put in motion move i of an inch the first second of 
time, 1 in. the second sec, 3 in. the third, and so continue to increase 
in geometrical ratio, how far would it move in 30 seconds ? 



MISCELLANEOUS EXAMPLES. 419 

67. If stock bought at 5 ^ premium will pay 6 ^ on the invest- 
ment, what °jo will it pay if bought at 15 <f discount ? 

68. If 6 apples and 7 peaches cost 33 cts., and 10 apples and 8 
peaches cost 44 cts., what is the price of one of each? 

Ans. Apples, 2 cts. ; peaches, 3 cts. 

69. A gentleman in dividing his estate among his sons gave A $9 
as often as B $5, and C $3 as often as B $7. C's share was $3862.50; 
what was the value of the whole estate ? 

70. A farmer sold 16 bu. of corn and 20 bu. of rye for $30, and 24 
bu. of corn and 10 bu. of rye for $27. How much per bushel did he 
receive for each ? 

71. A drover sold some oxen at $28, cows at $17, and sheep at 
$7.50 per head, and received $749 for the lot. There were twice as 
many cows as oxen, and three times as many sheep as cows. How 
many were there of each kind ? 

72. For what sum must a vessel, valued at $25000, be insured, so 
that in case of its loss, the owners may recover both the value of the 
vessel and the premium of 24 ^ ? 

73. A boy hired to a mechanic for 20 weeks, on condition that he 
should receive $20 and a coat. At the end of 12 weeks the boy quit 
work, when it was found that he was entitled to $9 and the coat ; 
what was the value of the coat ? 

74. An irregular piece of land, containing 540 A. 36 P., is ex- 
changed for a square piece containing the same area ; what is the 
length of one of its sides ? If divided into 42 equal squares, what 
will be the length of the side of each ? 

75. What] will be the difference in the expense of fencing two fields 
of 25 acres each, one in the form of a circle, and the other in the form 
of a square, the fence costing $.62J a rod? 

76. At what time between 5 and 6 o'clock are the hour and minute 
hands of a watch exactly together ? 

77. A general, forming his army into a square, had 284 men re- 
maining ; but increasing each side by one man, he wanted 25 men to 
complete the square. How many men had he ? Ans. 2400. 

78. Divide $3648 among 3 persons, so that the share of the first to 
that of the second shall be as 7 to 9, and of the first to the third as 3 
to 4. 

79. If a lot of land, in the form of an oblong or rectangle, contains 
7 A, 1 R. 10 P., and it requires 142 rd. of fence to inclose it, what 
are its dimensions ? Ans. 42 rd. by 26 rd. 

80. Five persons are employed to build a house. A, B, C, and D 
can build it in 13 days ; A, B, C, and E in 15 days ; A, B, D, and E 
in 12 days ; A, C, D, and E in 19 days ; and B, C, D, and E in 14 
days. In how many days can all together build it ; and should I de- 
sire to discharge 4 of them, which one would finish the work alone in 
the shortest time ? 

81. Divide $500 among 3 persons, in such a manner that the share 
of the second may be J greater than that of the first, and the share 
of the third J greater than that of the second. 

82. A gentleman who was entitled to a perpetuity of $3000 a year, 
provided in his will that, after his decease, his oldest son should receive 



120 MISCELLANEOUS EXAMPLES. q 

it for 10 yr., then his second son for the next 10 yr., and a literary 
institution for ever afterward. What was the value of each bequest ^\ 
at the time of his decease, allowing compound interest at 6 <f ? 

83. B has 3 teams engaged in transportation ; his horse team can 
perform the trip in 5 days, the mule team in 7 days, and the ox team 
m 11 days. Provided they start together, and each team rests a day 
after each trip, how many days will elapse before they all rest the 
same day ? 

84. A man bought a farm for $4500, and agreed to pay principal 
and interest in 4 equal annual instalments ; how much was the annual 
payment, interest being 6 % ? 

85. A bought a piece of property of B, and gave him his bond for 
$6300, dated Jan. 1, 1860, payable in 6 equal annual instalments of 
$1050, the first to be paid Jan. 1, 1861. A took up his bond Jan. 
1, 1864, semi-annual discount at the rate of 6 % per annum on the 
several payments which fell due after Jan. 1, 1861, being deducted ; 
what sum cancelled the bond ? 

86. A gentleman desires to set out an orchard of 864 trees, so that 
the length shall be to the breadth as 3 to 2, and the distance apart 
of each tree shall be 7 yards ; how many trees must there be in 
length, and how many square yards of ground will they occupy ? 

Arts. 36 in length ; they will occupy 39445 sq. yd. 

87. S. C. Wilder bought 25 shares of bank stock at an advance of 

6 % on the par value of $100. From the time of purchase until at 
the end of 3 yr. 3 mo. he received a semi-annual dividend of 4 % , 
when he sold the stock at a premium of 11 % . Money being worth 

7 ft compound interest, how much did he gain ? 

88. A builder employs a certain number of men, dividing them into 
companies according to the kind of work they can perform. When 
he settles with each company at the end of the week, he finds that the 
number of men in each is such, that they constitute a geometrical 
progression whose first term is 4, last term 64, and ratio 2. He set- 
tles with the men in such a way that, by paying $1.50 per day each 
to those in the largest company, and $.50 per day to those in the 
smallest company, and taking one day's wages off one man in each 
company, he has an arithmetical progression whose common differ- 
ence is .25. Find the number of companies ; the total number of 
men ; the daily wages in each company ; and the total sum paid for 
one week. 

89. If 20 sheep or 12 colts consume 4 acres of pasture in 6 weeks, 
and 30 sheep and 27 colts consume 18| acres in 8 weeks, and allow- 
ing that 45 sheep are equal to 18 horses, and that 10 horses with 12 
colts are equal to 9 oxen with 15 sheep; required the time it will take 
30 oxen and 100 sheep to consume a pasture of 80 acres, after 24 
horses and 30 colts have been pastured upon it 1 week and 1 day, 
the grass growing uniformly. 



Notice. — A Key to this work is published, containing full and clear solutions 
to all tho example? 



Oct 1863 



LIBRARY OF CONGRESS 








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